[asy][asy] unitsize(100); pen n_purple = rgb(0.7,0.4,1), n_blue = rgb(0,0.6,1), n_green = rgb(0,0.4,0), n_orange = rgb(1,0.4,0.1), n_red = rgb(1,0.2,0.4); pair A, B, C, D, H, M, N, S, T, E, M1, N1; A = dir(110); B = dir(-10); D = dir(190); H = foot(A, B, D); C = 2 * foot(origin, A, H) - A; M = (B + C)/2; N = (C + D)/2; S = foot(H, A, D); T = foot(H, A, B); E = 2 * foot(origin, A, extension(B, D, S, T)) - A; M1 = (B + H)/2; N1 = (D + H)/2; fill(E--C--D--cycle^^E--B--H--cycle, rgb(0.5,1,0.5) + opacity(0.5)); draw(A--C^^B--D, gray(0.5)); draw(M--S^^N--T, gray(0.7)); draw(circumcircle(A, S, T), n_purple); draw(circumcircle(E, B, H)^^circumcircle(E, D, H), n_red); draw(E--B^^E--D^^E--H^^E--C, n_green); draw(unitcircle, n_blue); draw(A--B--C--D--cycle); dot(A^^B^^C^^D^^H^^M^^N^^S^^T^^E^^M1^^N1); label("$A$", A, dir(100)); label("$B$", B, dir(310)); label("$C$", C, dir(C)); label("$D$", D, dir(220)); label("$H$", H, dir(290)); label("$M$", M, rotate(-90, origin) * dir(B - C)); label("$N$", N, rotate(90, origin) * dir(D - C)); label("$S$", S, dir(170)); label("$T$", T, dir(30)); label("$E$", E, dir(170)); label("$M_1$", M1, rotate(-90, origin) * dir(B - H)); label("$N_1$", N1, rotate(90, origin) * dir(D - H)); [/asy][/asy] Let $E$ be the second intersection of $\odot(ABD)$ and $\odot(AST)$. We contend that $E$ has the requested properties. First of all, since $\overline{AC} \perp \overline{BD}$ it follows easily that $\overline{HS} \perp \overline{AD}$ and $\overline{HT} \perp \overline{AB}$. It follows that $S$ and $T$ lie on $\odot(AST) = \odot(AH)$. To show $\angle BEH = \angle HES$, note that \[\angle BEH = 90^{\circ} - \angle BEA = 90^{\circ} - \angle BDA = \angle HAS = \angle HES\]as desired; other configurations are handled similarly. Similarly $\angle DEH = \angle HET$. Note $\angle BEH = 90^{\circ} - \angle BEA = 90^{\circ} - \angle BCH = \angle HBC$. It follows that $\overline{MB}$ is tangent to $\odot(ABH)$. Since $\angle BHC = 90^{\circ}$, $MB = MH$, so $\overline{MH}$ is also tangent to $\odot(ABH)$. Hence $\overline{EM}$ is the $E$-symmedian of $\triangle EBH$. Similarly $\overline{EN}$ is the $E$-symmedian of $\triangle EDH$. We claim $\triangle ECD \sim \triangle EBH$ now. Indeed $\angle ECD = \angle EBH$ and $\angle EDC = 180^{\circ} - \angle EHD = \angle EHB$, so $\triangle ECD \sim \triangle EBH$. Letting $M_1$ be the midpoint of $\overline{BH}$, note $ECND \sim EBM_1H$. Then \[\angle DEN = \angle HEM_1 = \angle BEM \implies \angle BEN = \angle MED\]as desired.

Probably similar to other solutions with 2011 G4, but whatever. Let the circle with diameter $\overline{AH}$ hit the circumcircle of $ABCD$ again at $E$. We claim this point $E$ satisfies both of the conditions. [asy][asy] size(250); defaultpen(fontsize(10pt)); pair A, B, C, D, H, S, T, M, N, E, X, Y; C = dir(110); B = dir(200); D = dir(340); H = foot(C, B, D); A = IP(Line(C, H, 10), circumcircle(B, C, D), 1); M = midpoint(B--C); N = midpoint(C--D); S = extension(M, H, A, D); T = extension(N, H, A, B); E = IP(circumcircle(A, S, T), circumcircle(A, B, C), 0); X = IP(Line(E, M, 10), circumcircle(B, C, D), 1); Y = IP(Line(E, N, 10), circumcircle(B, C, D), 1); draw(circumcircle(A, B, C), red); draw(A--B--C--D--cycle, orange); draw(A--C^^B--D, orange); draw(M--S^^N--T, heavygreen); draw(X--E--Y, magenta); draw(B--E--T^^D--E--S, blue); draw(circumcircle(A, S, T), purple); draw(circumcircle(M, N, E), heavycyan+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$M$", M, dir(140)); dot("$N$", N, dir(0)); dot("$S$", S, dir(300)); dot("$T$", T, dir(110)); dot("$E$", E, dir(E)); dot("$H$", H, dir(70)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); [/asy][/asy] First, observe that $S$ and $T$ lie on the circle with diameter $\overline{AH}$. Then, since $$\angle BEH = \angle BEA-90^\circ = \angle BDA+90^\circ = \angle CBD = \angle SAH = \angle HES,$$$\overline{EH}$ bisects $\angle BES$. We also know that $E$ is the center of the spiral similarity that sense $\overline{BT}$ to $\overline{DS}$, so $\angle BET = \angle DES$. Thus, $\overline{EH}$ also bisects $\angle TED$. Since $\angle BEH = \angle CBD$, we know that the circumcircle of $MNE$ is tangent to the circumcircle of $BCD$ by Shortlist 2011 G4. Thus if $X$, $Y$ are the second intersections of lines $EM$, $EN$ with the circumcircle of $BCD$, then $\overline{XY}\parallel \overline{MN}\parallel \overline{BD}$ by homothety at $E$. Thus, $\angle BEM = \angle BEN$, so $\angle BEN= \angle MED$, which completes the proof.

Sniped! CantonMathGuy wrote: Let $ABCD$ be a convex cyclic quadrilateral which is not a kite, but whose diagonals are perpendicular and meet at $H$. Denote by $M$ and $N$ the midpoints of $\overline{BC}$ and $\overline{CD}$. Rays $MH$ and $NH$ meet $\overline{AD}$ and $\overline{AB}$ at $S$ and $T$, respectively. Prove that there exists a point $E$, lying outside quadrilateral $ABCD$, such that ray $EH$ bisects both angles $\angle BES$, $\angle TED$, and $\angle BEN = \angle MED$. Proposed by Evan Chen Define $E$ as the 2011 G4 point and $G \overset{\text{def}}=\overline{BN} \cap \overline{DM}$ be the centroid in $\triangle BCD$. The scrolls foretold that $\overline{EA}, \overline{EG}$ are isogonal in angle $BED$, hence the isogonality lemma yields that $\overline{EM}, \overline{EN}$ are also isogonal in that angle. Proving part (b). For the trickier part (a), consider an inversion at $H$ that swaps $\{A, B\}$ with $\{C, D\}$. Let $BCFD$ be an isosceles trapezoid with $\overline{CF} \parallel \overline{BD}$ and $BC=FD$. Let $M^{*}, N^{*}$ be reflections of $H$ in $M, N$ respectively. Then $X$ inverts to $F$; $S$ to $M^{*}$ and $T$ to $N^{*}$. Consequently, $$\angle HEB=\angle HDF=\angle CBD=\angle HM^{*}F=\angle HES$$proving $\overline{EH}$ bisects angle $BES$. Likewise we can check $\overline{EH}$ bisects angle $TED$, proving part (a).

Solution from the packet: The main claim is that $E$ is the intersection of $(ABCD)$ with the circle with diameter $\overline{AH}$. [asy][asy] size(10cm); pair A = dir(100); pair B = dir(190); pair D = dir(-10); pair F = -A; pair H = foot(A, B, D); pair E = foot(A, F, H); pair C = -A+2*foot(origin, A, H); draw(unitcircle, lightblue); draw(A--B--C--D--cycle, lightblue); draw(A--B--F--D--cycle, lightblue); draw(A--C, lightblue); draw(B--D, lightblue); pair M = midpoint(B--C); pair N = midpoint(D--C); pair S = foot(H, A, D); pair T = foot(H, A, B); draw(M--S, lightblue); draw(N--T, lightblue); draw(circumcircle(B, T, S), orange); // pair Q = midpoint(A--H); draw(H--F, lightblue); pair P = midpoint(H--F); draw(B--T--S--D--cycle, orange+1); // draw(P--Q, blue); draw(A--F, blue); draw(circumcircle(E, B, M), lightgreen); draw(circumcircle(E, D, N), lightgreen); draw(H--E--A, orange); draw(circumcircle(A, T, S), lightgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$D$", D, dir(D)); dot("$F$", F, dir(F)); dot("$H$", H, dir(H)); dot("$E$", E, dir(E)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$S$", S, dir(S)); dot("$T$", T, dir(T)); // dot("$Q$", Q, dir(200)); dot("$P$", P, dir(P)); /* TSQ Source: !size(10cm); A = dir 100 B = dir 190 D = dir -10 F = -A H = foot A B D E = foot A F H C = -A+2*foot origin A H unitcircle 0.1 lightcyan / lightblue A--B--C--D--cycle lightblue A--B--F--D--cycle lightblue A--C lightblue B--D lightblue M = midpoint B--C N = midpoint D--C S = foot H A D T = foot H A B M--S lightblue N--T lightblue circumcircle B T S 0.1 yellow / orange Q = midpoint A--H R200 H--F lightblue P = midpoint H--F B--T--S--D--cycle 0.1 yellow / orange+1 P--Q blue A--F blue circumcircle E B M lightgreen circumcircle E D N lightgreen H--E--A orange circumcircle A T S lightgreen */ [/asy][/asy] The following observation can be quickly made without reference to $E$. Lemma: We have $\angle HSA = \angle HTA = 90^{\circ}$. Consequently, quadrilateral $BTSD$ is cyclic. Proof. This is direct angle chasing. In fact, $\overline{HM}$ passes through the circumcenter of $\triangle BHC$ and $\triangle HAD \sim \triangle HCB$, so $\overline{HS}$ ought to be the altitude of $\triangle HAD$. $\blacksquare$ From here it follows that $E$ is the Miquel point of cyclic quadrilateral $BTSD$. Define $F$ to be the point diametrically opposite $A$, so that $E$, $H$, $F$ are collinear, $\overline{CF} \parallel \overline{BD}$. By now we already have \[ \measuredangle BEH = \measuredangle BEF = \measuredangle BAF = \measuredangle CAD = \measuredangle HAS = \measuredangle HES \]so $\overline{EH}$ bisects $\angle BES$, and $\angle TED$. Hence it only remains to show $\angle BEM = \angle NED$; we present several proofs below. First proof (original solution) Let $P$ be the circumcenter of $BTSD$. The properties of the Miquel point imply $P$ lies on the common bisector $\overline{EH}$ already, and it also lien on the perpendicular bisector of $\overline{BD}$, hence it must be the midpoint of $\overline{HF}$. We now contend quadrilaterals $BMPS$ and $DNPT$ are cyclic. Obviously $\overline{MP}$ is the external angle bisector of $\angle BMS$, and $PB = PS$, so $P$ is the arc midpoint of $(BMS)$. The proof for $DNPT$ is analogous. It remains to show $\angle BEN = \angle MED$, or equivalently $\angle BEM = \angle NED$. By properties of Miquel point we have $E \in (BMPS) \cap (TPND)$, so \[ \measuredangle BEM = \measuredangle BPM = \measuredangle PBD = \measuredangle BDP = \measuredangle NPD = \measuredangle NED \]as desired. Second proof (2011 G4) By 2011 G4, the circumcircle of $\triangle EMN$ is tangent to the circumcircle of $ABCD$. Hence if we extend $\overline{EM}$ and $\overline{EN}$ to meet $(ABCD)$ again at $X$ and $Y$, we get $\overline{XY} \parallel \overline{MN} \parallel \overline{BD}$. Thus $\measuredangle BEM = \measuredangle BEX = \measuredangle YED = \measuredangle NED$. Third proof (Desargue involution, submitted by contestants) Let $G = \overline{BN} \cap \overline{MD}$ denote the centroid of $\triangle BCD$, and note that it lies on $\overline{EHF}$. Now consider the dual of Desargue involution theorem on complete quadrilateral $BNMDCG$ at point $E$; we get $(EB,EN)$, $(ED,EM)$, $(EC,EG)$ form an involutive pairing. However, the bisector of $\angle BED$, say $\ell$, is also the angle bisector of $\angle CEF$ (since $\overline{CF} \parallel \overline{BD}$) and so the involutions coincide with reflection across $\ell$. This means $\angle MEN$ is bisected by $\ell$ as well, as desired.

Nice problem! Don't have a diagram for my solution, so might be wrong somewhere. Let $E$ be the Miquel point of $STBD$. Firstly note that by Brahmagupta's theorem, $TEASH$ is cyclic. So angle chasing gives $\angle HED = 90^\circ - \angle DEA = \angle CDB = \angle TAH = \angle TEH$, which by symmetry verifies the first property. ISL 2011 G4 gives $(MEN)$ is tangent to $(ABCD)$. So the homothety at $E$ maps $(MEN)$ to $(ABCD)$, so we have, by equality of arcs, $\angle BEM = \angle NED$, which gives the second condition.

Let $\Gamma$ be the circumcircle, $X\in \Gamma$ with $CX\parallel BD$, and define $\{X,E'\}\equiv XH\cap \Gamma$. Pentagon $THSAE'$ is cyclic, so $\angle BE'H=\angle BDX=\angle CBD=\angle SAC=\angle SE'H$. It is well-known (say, by homothety), that the centroid $G$ of $\triangle BCD$ lies on $\overline{HX}$; moreover, since $\angle BE'C=\angle BDC=\angle XBC=\angle XE'C$, we know that $\overline{E'C}$ and $\overline{E'G}$ are isogonal in $\angle BE'C$. Now by dual of Desargues' involution theorem on complete quadrilateral $BNDMGC$, the involution that swaps $\{E'B, E'D\}$ and $\{E'C, E'G\}$ also swaps $\{E'M, E'N\}$, which implies the isogonality; hence, $E'\equiv E$.

First of all let's give some names, let $DM\cap BN=K$ and $BS \cap DT=Z$. Now the basics: due to the remarkable result of Brahmagupta we can easily deduce that $DSTB$ is cyclic and we have that $\angle{HTA}=\angle{HSA}=90^\circ$. Converse of Pascal on $TBMSDN$ and note real quick that $T, B, M, S, D, N$ are on a conic, now Pascal on $SBNTDM$ and we see that $Z, H, K$ lie on a line. Now add $J$, the Miquel point of $DSTB$, now because $\angle{HJA}=\angle{HTA}$, due to 2 well known proprieties we have that $HJ$ passes through the circumcenter of $DSTB$ (thus we are done with a.), also it passes through $Z$, thus it passes through $K$. Note that $\angle{JCD}=\angle{JBD}$ and that $\angle{BHJ}=\angle{HAJ}=\angle{CDB}$ thus $\angle{DJK}=\angle{BJC}$. Now by isogonality lemma on $\{K,C\}$ wrt $\angle{BJD}$ we see that $\angle BJN = \angle MJD$, now using one more well-known propriety we actually see that $J$ is our wanted $E$, thus we are done. The "well-know" proprieties used: In a cyclic quad ABCD we have the following things: 1) the intersection of diagonals is the inverse of Miquel point (call it M) 2) Miquel point lies on the "outer diagonal??" 3) if O is the circumcenter of ABCD, then MO bisect angles BMD and AMC.

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@v_Enhance How did you come up with this nice problem?

Some motivation on finding the point $E$: An effective method is to invert at $H$, which greatly simplifies the diagram. Then using the condition that $EH$ bisects $\angle BES$ and $\angle TED$, we can find an explicit construction for $E.$ As discussed in the solutions above, using the fact that $M$ and $N$ are the circumcenters of $\triangle HBC$ and $\triangle HDC$, we can readily find that $HS \perp AD$ and $HT \perp AB.$ In particular, $A, H, S, T$ are inscribed in the circle of diameter $\overline{AH}.$ In addition, since $H$ lies on the $A$-altitude of $\triangle ABD$, it follows that $ST$ is antiparallel to $BD$, and hence $BDST$ is cyclic. Now, invert at $H$ and denote inverses with primes. We see that $B'D'S'T'$ is cyclic, and the angle condition $\angle HB'T' = \angle HD'S' = 90^{\circ}$ forces it to be a rectangle. Moreover, $A', S', T'$ are collinear, and $H'A' \perp B'D'.$ Thus, we obtain a very simple picture: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 978.0355033181172, xmax = 1050.9089588785569, ymin = 1016.2406481893914, ymax = 1032.275243882142; /* image dimensions */ pen fsfsfs = rgb(0.9490196078431372,0.9490196078431372,0.9490196078431372); filldraw((998.7640266232867,1028.1588688773386)--(998.6643819711985,1018.3272632046446)--(1025.7350980984547,1018.0528978384899)--(1025.834742750543,1027.8845035111838)--cycle, fsfsfs, linewidth(0.8) + gray); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((998.7640266232867,1028.1588688773386)--(998.6643819711985,1018.3272632046446), linewidth(0.8) + gray); draw((998.6643819711985,1018.3272632046446)--(1025.7350980984547,1018.0528978384899), linewidth(0.8) + gray); draw((1025.7350980984547,1018.0528978384899)--(1025.834742750543,1027.8845035111838), linewidth(0.8) + gray); draw((1025.834742750543,1027.8845035111838)--(998.7640266232867,1028.1588688773386), linewidth(0.8) + gray); /* dots and labels */ dot((998.7640266232867,1028.1588688773386),dotstyle); label("$T'$", (997.6654997794753,1028.5468363406546), NE * labelscalefactor); dot((998.6643819711985,1018.3272632046446),dotstyle); label("$B'$", (997.7983593155928,1017.2205608866386), N * labelscalefactor); dot((1025.7350980984547,1018.0528978384899),dotstyle); label("$D'$", (1026.031010740558,1016.8551971623154), NE * labelscalefactor); dot((1025.834742750543,1027.8845035111838),dotstyle); label("$S'$", (1025.9645809724993,1028.214687500361), NE * labelscalefactor); dot((1009.5909847616014,1018.2165206087959),dotstyle); label("$H$", (1009.2574943057258,1016.988056698433), N * labelscalefactor); dot((1009.6906294136895,1028.04812628149),dotstyle); label("$A'$", (1009.4235687258727,1028.746125644831), NE * labelscalefactor); dot((1014.90813996014,1027.9952461070327),dotstyle); label("$E'$", (1015.0368841268363,1028.314332152449), NE * labelscalefactor); dot((1012.2495623608706,1023.1058833579143),dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] After inversion, the angle bisector condition becomes $\angle HB'E' = \angle HS'E'$ and $\angle HT'E' = \angle HD'E'.$ Inspired by the great symmetry of the figure, it's not too hard to guess that $E'$ should be the point on the rectangle symmetric to $H.$ In particular, $E' \in \overline{A'S'T'}$ and $A', B', D', E'$ are concyclic by symmetry. These observations show that back in the original figure, $E$ is the second intersection of $\odot(AHST)$ and $\odot(ABD).$

pinetree1 wrote: Probably similar to other solutions with 2011 G4, but whatever. Let the circle with diameter $\overline{AH}$ hit the circumcircle of $ABCD$ again at $E$. We claim this point $E$ satisfies both of the conditions. [asy][asy] size(250); defaultpen(fontsize(10pt)); pair A, B, C, D, H, S, T, M, N, E, X, Y; C = dir(110); B = dir(200); D = dir(340); H = foot(C, B, D); A = IP(Line(C, H, 10), circumcircle(B, C, D), 1); M = midpoint(B--C); N = midpoint(C--D); S = extension(M, H, A, D); T = extension(N, H, A, B); E = IP(circumcircle(A, S, T), circumcircle(A, B, C), 0); X = IP(Line(E, M, 10), circumcircle(B, C, D), 1); Y = IP(Line(E, N, 10), circumcircle(B, C, D), 1); draw(circumcircle(A, B, C), red); draw(A--B--C--D--cycle, orange); draw(A--C^^B--D, orange); draw(M--S^^N--T, heavygreen); draw(X--E--Y, magenta); draw(B--E--T^^D--E--S, blue); draw(circumcircle(A, S, T), purple); draw(circumcircle(M, N, E), heavycyan+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$M$", M, dir(140)); dot("$N$", N, dir(0)); dot("$S$", S, dir(300)); dot("$T$", T, dir(110)); dot("$E$", E, dir(E)); dot("$H$", H, dir(70)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); [/asy][/asy] First, observe that $S$ and $T$ lie on the circle with diameter $\overline{AH}$. Then, since $$\angle BEH = \angle BEA-90^\circ = \angle BDA+90^\circ = \angle CBD = \angle SAH = \angle HES,$$$\overline{EH}$ bisects $\angle BES$. We also know that $E$ is the center of the spiral similarity that sense $\overline{BT}$ to $\overline{DS}$, so $\angle BET = \angle DES$. Thus, $\overline{EH}$ also bisects $\angle TED$. Since $\angle BEH = \angle CBD$, we know that the circumcircle of $MNE$ is tangent to the circumcircle of $BCD$ by Shortlist 2011 G4. Thus if $X$, $Y$ are the second intersections of lines $EM$, $EN$ with the circumcircle of $BCD$, then $\overline{XY}\parallel \overline{MN}\parallel \overline{BD}$ by homothety at $E$. Thus, $\angle BEM = \angle BEN$, so $\angle BEN= \angle MED$, which completes the proof. Nice solution. (Better than mine, as I bash the second part). Note that it is sufficient to show that $\angle BEM = \angle NED .$ This is equivalent to proving that $cot {\angle BEM}=\cot{ \angle NED},$ which after computations becomes $\cot{ \angle BAH} - \cot{ \angle DAH} = \cot{ \angle EAH}.$ Intersecting $EA$ and $BD$ and using power of a point gives the result.

CantonMathGuy, Restated in terms of triangle BCD wrote: Let $H$ be the foot of the $A$-altitude in $\triangle ABC$, and let $AH \cap \omega=D$, where $\omega$ is the circumcircle of $\triangle ABC$. Denote by $M$ and $N$ the midpoints of $\overline{AB}$ and $\overline{AC}$. Rays $MH$ and $NH$ meet $\overline{CD}$ and $\overline{BD}$ at $S$ and $T$ respectively. Prove that there exists a point $E$, lying outside quadrilateral $ABCD$, such that ray $EH$ bisects both angles $\angle BES$, $\angle CET$, and $\angle BEM = \angle CEN$. Proposed by Evan Chen Here's my solution (Probably a bit different): WLOG assume that $AB>AC$. Let $G$ be the centroid of $\triangle ABC$, and $GH \cap \omega =E,F$, with $E$ lying on arc $\overarc{BC}$ not containing $A$. We claim that this point $E$ satisfies both the given properties. In the second property, we wish to prove that $EM$ and $EN$ are isogonal in $\angle BEC$. By the Isogonality Lemma, it suffices to show that $EG$ and $EA$ are isogonal in $\angle BEC$. Now consider the homothety $\mathcal{H}(G,-2)$, which takes $H$ to the point $F'$ on $\omega$ such that $AF'BC$ is an isosceles trapezoid. This gives $F'=F$, and so $\angle BEG=\angle BAF=\angle CBA=\angle CEA$, showing that $E$ satisfies Property 2. Now, it's evident that $F$ is the antipode of $D$ in $\omega$, and so $\angle DEH=90^{\circ}$, which means that $E$ lies on $\odot (DH)$. Also, if $P$ is the midpoint of the segment joining $C$ and the orthocenter of $\triangle ABC$, and $K$ is the foot of the $B$-altitude, then it is well known that $MH$ is tangent to $\odot (CHK)$, which has center $P$. But, $\angle CHP=90^{\circ}-\angle CKH=90^{\circ}-B=\angle BAD=\angle HCD$. All these statements together imply that $MH \perp HP \parallel CD \Rightarrow S$ lies on $\odot (DH)$. Similarly, $T$ also lies on this circle. Now, Let $Q$ be the point on $\omega$ such that $AFCQ$ is an isosceles trapezoid. Then $\angle FEQ=\angle FCQ=\angle AFC=\angle ABC$ and $\angle FES=\angle HES=\angle HDS=\angle ABC$. Thus, we get that $E,S,Q$ are collinear. But, $FQ=AD=BF$ implies that $F$ is the midpoint of arc $\overarc{BQ}$, giving that $EH$ bisects $\angle BES$. Taking analogous result for $\angle DET$, we have that $E$ satisfies Property 1 also. Hence, done. $\blacksquare$

We get $ATHS$ cyclic. Let $Q=(ATS)\cap BE$, $E=(AB)\cap (ATS)$ and $R=(ATS)\cap DE$ Applying Pascal's theorem on $AASTRE$ and $ATSQEA$ we get $BD||QS||TR$ Which gives that $EH$ is the angle bisector of $\angle{BES}$ and $\angle{TED}$ Then angle chase gives $\angle{ETN}+\angle{EDN}=\angle{ETS}+\angle{HTS}+\angle{EDN}=\angle{EQS}+\angle{CBD}+\angle{EDN}=\angle{EBD}+\angle{CBD}+\angle{EDN}=180$ So, $ETND$ is cyclic. Similarly $EBMS$ will be cyclic. At last, $AH^2=AT.AB=AS.AD$ So, $BTSD$ is cyclic. $\angle{BTD}=\angle{BSD}$ So, $\angle{BSM}=\angle{NTD}$ as $\angle{BTH}=\angle{DSH}=90$ So, $\angle{BEM}=\angle{NED}$ implies $\angle{BEN}=\angle{MED}$ Which means $E$ is the desired point.

Let $\angle MCH = x, \angle NCH = y$. Easy angle chasing gives $SATH$ is cyclic with diameter $AH$. This gives $\angle AST = y = \angle ABD$, so $STBD$ is cyclic, let's say with circumcenter $O$ and Miquel point $E$. We claim this choice of $E$ works. $E$ lies on $(ABC)$, so it is outside $ABCD$. Note $E$ lies on $(AH)$, so $E, H, O$ are collinear. It is well-known that $EO$ bisects $\angle BES$ and $\angle TED$ (easy proof: invert about $(STBD)$). Let $EH$ meet $(ABCD)$ again at $L$. $\angle BEL = \angle HES = \angle CAD$, and $L$ and $C$ lies on the same side of $BD$, so $BCLD$ is an isosceles trapezium with $BD \parallel CL$. Since $H$ is the foot of the $C-$altitude in $CBD$, if $G$ is the centroid of $CBD$, then $G, H, L$ are collinear (easy proof: homothety at $G$ with ratio $-2$). This implies $EG$ and $EC$ are isogonal in $\angle BED$. By the isogonal line lemma and the fact that $BG \cap DC = N, DG \cap BC = M$, $EM$ and $EN$ are isogonal in $\angle BED$, and we're done.

Let $E$ be the second intersection of $(ATS)$ and $(ABCD)$, we will show that $E$ is the desired point. Firstly, notice that $\angle BHT=\angle NHD=\angle HDN=\angle =\angle CBD=\angle BAC$, hence $HT\perp AB$ and similarly $HS\perp AD$. THereore $H$ also lies on $(TSH)$. Now we have $\angle TSH=\angle TAH=\angle BDC$ and therefore $\triangle TSH\sim\triangle BDC$. By sprial similarity lemma, $E$ is the center of spiral similarity sending $\triangle TSH$ to $\triangle BDC$. Now let $ED$ meet $(ATS)$ again at $P$, then $\angle EPT=\angle EST=\angle EDB$ which implies $TP\|BD$, since $BD$ is tangent to the circle $(ATS)$ at $H$, $HT=HP$. Hence $\angle TEH=\angle PEH=\angle DET$ as desired. Similarly $\angle BEH=\angle SEH$. Now let $G$ be the centroid of $BCD$. Let $A'$ be the reflection of $A$ in the circumcenter of $(ABC)$, then since $\angle AEA'=\angle AEH=90^{\circ}$, $E,H,A'$ are collinear. Notice that using complex numbers with $(ABCD)$ as the unit circle we easily obtain that $$3(g-a')=2(h-g)=3a+b+d-\frac{bd}{a}$$hence $H,G,E$ are collinear. Therefore $\angle BEC=\angle BAC=\angle BEH=\angle HED=\angle GED$. Therefore by applying isogonal lemma on $\angle BED$ and $\angle CEN$, we have $\angle BEM=\angle DEN$ as desired.

We will show that $E=(AH)\cap (ABC)$ satisfies all 3 properties listed. First, we show that $\angle BEN=\angle MED$. We want to show that $EM$ and $EN$ are isogonal in $\angle BED$. Let $C'$ be the point such that $CBDC'$ is an isosceles trapezoid, and let $\ell$ be line $HG$. We claim $C'\in \ell$. Indeed, if we let $L_1$ be the midpoint of $BD$, then taking a homothety at $G$ with scale $-2$ sends $L_1NM\to CBD$, so reflecting over the perpendicular bisector of $BD$ gives that it also sends $H \to C'$, proving $C'\in \ell$. We claim also $E\in \ell$. Note that $C'$ is the antipode of $A$, since $O$ lies on the perpendicular bisectors of both $CC'$ and $CA$; this means $\angle C'EA=90$. Since $\angle HEA=90$ as well, this implies $E\in HC' = \ell$. Now, there are two finishes: By the Second Isogonality Lemma, it suffices to show the lines through $E$ and (1) $BM\cap DN=C$, (2) $BN\cap MD=G$, the centroid of $\triangle BCD$, are isogonal in $\angle BED$. But these are simply lines $EC$ and $EC'$. Since the arc midpoint of $CC'$ coincides with the arc midpoint of $\widehat{BCD}$, this is clear. $~$ Since $E=HG \cap (ABCD)$, the converse of ISL 2011 G4 implies $(EMN)$ is tangent to $(ABCD)$ at $E$; hence $EM$ and $EN$ are isogonal in $\angle BED$. Now, we show $EH$ bisects $\angle BES$ (the other follows by symmetry). First we claim $S,T \in (AH)$. Indeed, \begin{align*} \angle SHA &= \angle CHM = \angle HCM = \angle ACB=\angle ADB=90-\angle HAD, \end{align*}since $M$ is the circumcenter of right triangle $\triangle CHB$. Similarly, $T\in (AH)$. Now, \begin{align*} \angle BEH &= \angle BEC' = \angle BDC' = \angle DBC\\ \angle SEH &= \angle SAH = \angle DAC=\angle DBC, \end{align*}and we are done.

Let $E = (AH) \cap (ABC)$. Claim: Ray $EH$ bisects both angles $\angle BES$, $\angle TED$. Proof: I first claim that $S,T \in (AH)$. Indeed, one can simply remark by angle chasing that $\triangle BTH \sim \triangle CHD$, so $\angle BTH = \angle CHD = 90^{\circ}$. Now $\angle BEH = 90^{\circ} - \angle BEA = 90^{\circ} - \angle BDA = \angle DAH = \angle SEH$, thus ray $EH$ bisects $\angle BES$. Similarly it also bisects $\angle TED$. Claim: $\angle BEN = \angle MED$. Let $EH$ meet $(ABCD)$ again at $Y$; note that $Y$ is the antipode of $A$ and hence $BYCD$ is an isosceles trapezoid. But let $G$ be the centroid of $\triangle BCD$, by homothety $\overline{HGY}$ is collinear, this implies $\overline{EHG}$ collinear. Now by the converse of 2011 G4, we find that $(MNE)$ and $(ABCD)$ are tangent. This implies that $EM$ and $EN$ are isogonal in $\triangle EBD$, as desired.

I claim the point $E$ is in fact $(AH) \cap (ABC)$; redefine it as such. Note $T,S \in (AH)$ (say, because median and altitude are isogonal in right triangles). Note $\angle TEH = \angle TAH = 90^\circ - \angle ABD = 90^\circ - \angle AED = \angle DEH$ so $EH$ bisects angle $\angle TED$. Similarly it also bisects $\angle SEB$, so the first property is satisfied. Now let $EH$ intersect $(ABCD)$ again at $Y$, and let the centroid of $\Delta BCD$ be $X$. Since $\angle AEY = \angle AEH = 90^\circ$ we have $CY \parallel BD$, and thus $X \in \overline{EHY}$ by a well-known lemma. Finally, since $EC, EX$ are isogonal in angle $\angle BED$, we conclude by isogonality lemma that $M=DX \cap BC, N=BX \cap DC$ are also isogonal in angle $\angle BED$ as desired.

[asy][asy] size(10cm); defaultpen(fontsize(9pt)); defaultpen(linewidth(0.4)); dotfactor *= 1.5; pair A = dir(100), B = dir(210), D = dir(330), H = foot(A,B,D), C = H+dir(A--H)*abs(B-H)*abs(D-H)/abs(A-H), M = (B+C)/2, N = (C+D)/2, T = extension(A,B,H,N), S = extension(A,D,M,H), E = intersectionpoints(circumcircle(A,T,S),unitcircle)[1], P = (T+H)/2, Q = (S+H)/2, P1 = P+dir(E--P)*abs(T-P)*abs(H-P)/abs(E-P), Q1 = Q+dir(E--Q)*abs(H-Q)*abs(S-Q)/abs(E-Q); filldraw(E--M--P--cycle^^E--N--Q--cycle, cyan+white+white+white); draw(B--C--D^^A--C^^T--N--M--S); draw(A--B--D--A, linewidth(1)); draw(unitcircle, heavygreen); draw(circumcircle(A,T,S), orange); draw(P1--E--Q1^^P--M--E--N--Q, blue); draw(T--S^^P1--Q1^^P--Q, red); draw(circumcircle(P,M,N), magenta); draw(T--E--H^^E--D, purple+linewidth(0.6)); dot("$A$", A, dir(90)); dot("$B$", B, dir(210)); dot("$C$", C, dir(240)); dot("$D$", D, dir(330)); dot("$H$", H, dir(270)); dot("$M$", M, dir(225)); dot("$N$", N, dir(300)); dot("$T$", T, dir(180)); dot("$S$", S, dir(10)); dot("$E$", E, dir(180)); dot("$P$", P, dir(90)); dot("$Q$", Q, dir(90)); dot("$P_1$", P1, dir(270)); dot("$Q_1$", Q1, dir(315)); [/asy][/asy] Let $E = (AH) \cap (ABCD)$, $P$ be the midpoint of $\overline{HT}$, $Q$ be the midpoint of $\overline{HS}$, $P_1 = \overline{EP} \cap (AH)$, and $Q_1 = \overline{EQ} \cap (AH)$. I claim that $E$ is the desired point. Claim: $T, S \in (AH)$. Proof. We have $$\angle AHT = \angle CHN = \angle HCN = 90^\circ - \angle CDB = 90^\circ - \angle BAC,$$so $\overline{HT} \perp \overline{AB}$ and similarly $\overline{HS} \perp \overline{AD}$. $\square$ Claim: Ray $\overline{EH}$ bisects $\angle TED$ and $\angle BES$. Proof. We have $$\angle HED = 180^\circ - \angle EDH - \angle DHE = 180^\circ - \angle EAT - \angle HTE = \angle TEH,$$so $\overline{EH}$ bisects $\angle TED$ and similar for $\angle BES$. $\square$ Claim: $MNQP$ is cyclic. Proof. $\angle PNM = \angle BDC = \angle BAH = \angle TSH = \angle PQM$. $\square$ Claim: $\angle BEN = \angle DEM$, which finishes the problem. Proof. Remark that $\triangle EMP \sim \triangle ENQ$ by spiral similarity since $M$, $N$, $P$, $Q$ are midpoints. Then by Power of a Point \begin{align*} \frac{PP_1}{QQ_1} &= \frac{(HP \cdot PT) / EP}{(HQ \cdot QS) / EQ} \\ &= \frac{EQ}{EP} \cdot \left(\frac{HP}{HQ}\right)^2 \\ &= \frac{EQ}{EP} \cdot \left(\frac{MP}{NQ}\right)^2 \\ &= \frac{EQ}{EP} \cdot \left(\frac{EP}{EQ}\right)^2 \\ &= \frac{EP}{EQ},\end{align*}which implies $\overline{PQ} \parallel \overline{P_1Q_1}$. It follows (again by spiral similarity) that $$\angle BEN = \angle TEQ = \angle SEP = \angle DEM$$as desired. $\blacksquare$

First, we note that because $\angle NHD = \angle HDN = \angle BDC$, then we get $$180^{\circ} - \angle HTA = \angle TAH + \angle AHT = \angle BAC + \angle CHN = \angle BDC + 90^{\circ} - \angle NHD = 90^{\circ}$$and so $\angle HTA = 90^{\circ}$. Let $E$ be where the circle with diameter $AH$ hits the circumcircle of $\triangle ABC$. Since $\angle HTA = \angle HEA = \angle ASH$, then $AETHS$ is a cyclic pentagon. Since $BT$, $CH$, and $DS$ intersect at $A$, then a spiral similarity centered at $E$ takes $\triangle THS$ to $\triangle BCD$. Letting $A'$ be the antipode of $A$ with respect to the circumcircle of $\triangle ABC$. Then, $\angle AEH = \angle AEA' = 90^{\circ}$ which means that $E$, $H$, and $A'$ are collinear. In addition, note that $BD$ and $CA'$ are parallel. To prove that $\angle BEH = \angle HES$ and $\angle TEH = \angle HED$, we consider that $\angle BEH = \angle BEA' = \angle CED = \angle HES$ and $\angle TEH = \angle BEC = \angle A'ED = \angle HED$ from the spiral similarities above. As a result, $E$ satisfies the first condition. Note that $$\angle BHM = \angle MBH = \angle CBD = \angle CED = \angle BEA' = \angle BEH$$which means that the circumcircle of $\triangle EBH$ is tangent to $MH$ and $MB$, hence, letting $M'$ be the other intersection point of $EM$ and the circumcircle of $\triangle EBH$, we see that $(EM'; BH) = -1$ meaning that $EM'$ is a symmedian of $\angle EBH$. Let $M''$ be the midpoint of $BH$, so thus we see since $\angle BEH = \angle CED$ from before and $\angle HBE = \angle DBE = \angle DCE$, then a spiral similarity centered at $E$ takes $BH$ to $CD$, and thus takes $M''$ to $N$. Now, $\angle BEM = \angle M''EH = \angle NED$ which means that $$\angle BEM + \angle MEN = \angle NED + \angle MEN \to \angle BEN = \angle MED$$and we are done since we proved there exists such a point $E$ satisfying both conditions.

Beautiful problem. The solution below is just based on EGMO chapter 1 results and assumption of the properties of right triangles . Let $\gamma$ denote $(AST)$ and $\omega$ denote $(ABCD)$. We will show that $E=\gamma\cap\omega\ne A$ is the desired point. First, observe that \[\measuredangle ASH=\measuredangle AHS+\measuredangle SAH=\measuredangle CHM+\measuredangle DAC=90^\circ-\measuredangle HBC+\measuredangle DBC=90^\circ.\]Since we similarly have $\measuredangle ATH=90^\circ$, $AH$ is a diameter of $\gamma$. Observe that \[\measuredangle HBE=\measuredangle DBE=\measuredangle DAE=\measuredangle SAE=\measuredangle SHE,\]so line $SH$ is tangent to $(BEH)$. Then, check that \[\measuredangle BEH=\measuredangle BHM=\measuredangle HBM=\measuredangle DBC=\measuredangle DAC=\measuredangle SAH=\measuredangle SEH,\]so line $EH$ bisects $\angle BES$. Similarly, line $EH$ bisects $\angle TED$, so it remains to show $\angle BEN=\angle MED$. By the previous angle chase, we have \[\measuredangle BES=2\measuredangle BEH=2\measuredangle BHM=\measuredangle BHM+\measuredangle HBM=\measuredangle HMB=\measuredangle SMB,\]so $BESM$ is a cyclic quadrilateral. We have \[\measuredangle MED=\measuredangle MEB+\measuredangle BED= \measuredangle MSB+\measuredangle BAD=\measuredangle HSB+\measuredangle BAD.\]Similarly, we have $\measuredangle BEN = \measuredangle DTH+\measuredangle BAD$, so it suffices to show $\measuredangle DTH=\measuredangle HSB$. Now, remark that \[AT\cdot AB = AH^2 = AS\cdot AD\]by right triangle ratios, so $STBD$ is a cyclic quadrilateral by power of a point. This implies that \[\measuredangle HSB = \measuredangle DSB-90^\circ=\measuredangle DTB-90^\circ=\measuredangle DTH,\]as desired.

$\textbf{Claim:}$ $\overline{MS} \perp \overline{AD}$ and $\overline{NT} \perp \overline{AB}$ Well known, but included for completeness. We will use the fact that $M$ is the circumcenter of $\triangle BHC$, then just angle chase: \begin{align*} \measuredangle SAH + \measuredangle AHS = \measuredangle DBC + \measuredangle CHM = \measuredangle HBC + \measuredangle MCH = 90^\circ, \end{align*}thus done by symmetry. $\square$ $\textbf{Claim:}$ Let $E = (AHST) \cap (ABCD)$, then $E$ is the desired point. The first part is just angle chase. Noting the spiralsim at $E$ sending $\triangle EST$ to $\triangle EDB$, we have \begin{align*} \angle BEH = 180 - \angle EBD - \angle EHB = \angle EHD - \angle EBD = \angle ETH - \angle ETS = \angle STH = \angle SEH, \end{align*}as desired. Second part is a little more tricky; add in $C' = \overline{EH} \cap (ABCD)$, and add in the centroid $G$ of $\triangle CBD$. Note $\angle AEC' = \angle AEH = \angle ASH = 90^\circ$, which combined with $\angle HCC' = \angle ACC' = \angle AEC'$ yields $C'$ the point forming isosceles trapezoid $CC'BD$. Notice by nine point circle homothety that $H$, $G$, and $C'$ are collinear. Now, add in $(EMN)$, which by 2011 G4 is tangent to $(ABCD)$ at $E$. Now letting $N'$ and $M'$ be the images of homothety at $E$ sending $(ENM)$ to $(ABC)$, we find $\overline{M'N'} \parallel \overline{BD}$ which implies the desired isogonality.

We claim that $E$ is the IMO SL 2011 G4 point. Indeed, let $G= DM \cap BN \implies$ by homothety, $GH$ intersects $\widehat{BCD}$ on a point $H'$ such that $CH' \parallel BD$. $\implies$ let $E= GH \cap (ABCD), E \neq H' \implies EC,EH'$ are isogonal WRT $\angle BED$. $(*)$ By DDIT on $MBND$ WRT $E$, there is an involution swapping $(EM,EN),(EB,ED),(EC,EH) \implies$ from $(*)$, this involution is a reflection WRT the internal angle bisector of $\angle BED \implies EM,EN$ are isogonal WRT $\angle BED \implies \angle BEN= \angle MED \implies$ the second condition is satisfied. Now, observe that since $H'$ is the antipode of $A$ WRT $(ABCD)$, $\angle HEA= \angle H'EA= 90º$. Furthermore, since $HBC \sim HAD$ and $HM$ is $H$-median WRT $HBC \implies HS$ is $H$-symmedian of the right angled triangle $HAD \implies HS \perp AD$. Similarly, $HT \perp AB$, and since $\angle HEA=90º \implies (ETHSA)$ is cyclic. Thus, $\angle HES= \angle HAD= \angle CBD= \angle BDH'=\angle BEH$ ($BCH'D$ is an isosceles trapezoid), and similarly $\angle TEH= \angle DEH$, so ray $EH$ bisects both angles $\angle BES$ and $\angle TED$, as desired. $\blacksquare$

Below is a very fast proof of the first part by using some well known properties of Miquel point and isogonal lines (we will not even angle chase!): [asy][asy] size(200); pair A=dir(110),C=dir(-110),B=dir(0),D=dir(180),H=extension(A,C,B,D),M=1/2*(B+C),N=1/2*(D+C),S=foot(H,A,D),T=foot(H,A,B),E=2*foot(A,circumcenter(A,S,T),(0,0))-A,Ap=-A,Op=1/2*(H+Ap); filldraw(unitcircle,cyan); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$M$",M,dir(M)); dot("$N$",N,dir(-110)); dot("$E$",E,dir(E)); dot("$S$",S,dir(S)); dot("$T$",T,dir(T)); dot("$H$",H,dir(H)); dot("$A'$",Ap,dir(Ap)); dot("$O'$",Op,dir(-120)); draw(A--D--C--B--A^^A--C^^B--D^^D--Ap--B,magenta); draw(Ap--E,red+linewidth(0.9)); draw(M--S^^T--N,magenta); draw(circumcircle(D,Op,T)^^circumcircle(B,Op,S),green+linewidth(0.8)); draw(A--E,magenta); markscalefactor=0.01; draw(rightanglemark(A,E,Ap),darkblue); draw(rightanglemark(Ap,D,A),darkblue); draw(rightanglemark(Ap,B,A),darkblue); draw(rightanglemark(H,S,A),darkblue); draw(rightanglemark(H,T,A),darkblue); draw(rightanglemark(A,H,B),darkblue); [/asy][/asy] Set $E$ as the Miquel point of lines $DS,ST,TB,BD$. Let $A'$ be antipode of $A$ and $O'$ be midpoint of segment $A'H$. Note that $\overline{AH},\overline{AA'}$ are isogonal wrt $\angle BAD$ so points $D,S,T,B$ lie on a circle with center $O'$. It is well known that $\angle O'EA = 90^\circ$ and we also know (by definition) that $\angle AEA' = 90^\circ$. This is enough to imply points $E,H,O',A'$ are collinear. As lines $EH,EO'$ coincide, so it suffices to show line $EO'$ bisects $\angle BES$ (other result follows analogously). But is well known that points $E,S,O',B$ lie on a circle and since $\overline{O'B},\overline{O'S}$ are equal chords of this circle, we are done! $\square$

Solved with sriraamster. First of all, since $HM$ and the $H$-altitude to $BC$ are isogonal in $\angle HBC$ and lines $BC$ and $AD$ are antiparallel in the same angle, this implies that $HS$ is the $H$-altitude to $AD.$ Thus, $ASHT$ is cyclic with diameter $AH.$ This means that \[\measuredangle DST = 90^\circ + \measuredangle HST = 90^\circ + \measuredangle HAT = 90^\circ + (90^\circ - \measuredangle ABH) = \measuredangle DBT,\]so $DBTS$ is cyclic as well. Now let $E=(ABCD) \cap ATHS)\neq A$ and $O$ be the circumcenter of $DBTS.$ Claim: $DETON$ is cyclic. Proof: First, observe that \[\measuredangle TOD = 2(180^\circ - \measuredangle DST) = 2(180^\circ - (180^\circ - TSA)) = 2\measuredangle TSA = 2\measuredangle THA = 2\measuredangle ABD.\]Next, we can find that \[\measuredangle TND = \measuredangle NHC + \measuredangle HCN = \measuredangle THA + \measuredangle ACD = 2\measuredangle ABD.\]Finally, \[\measuredangle TED = \measuredangle TEA - \measuredangle DEA = (180^\circ - \measuredangle AHT) - \measuredangle DBA = 180^\circ - 2\measuredangle DBA = 2\measuredangle ABD,\]proving the claim. $\Box$ Back to the main problem: By symmetry our Claim also implies that $BESOM$ is cyclic. Furthermore, we also get that \begin{align*} \measuredangle CNO &=\measuredangle DNO =\measuredangle DTO \\ &=180^\circ - \measuredangle OTB - \measuredangle ATD \\ &=180^\circ - (90^\circ - \measuredangle BDT) - \measuredangle ATS - \measuredangle STD \\ &=90^\circ + \measuredangle BDT - \measuredangle BDS - \measuredangle SBD \\ &=90^\circ + \measuredangle BST + (\measuredangle DSB - 180^\circ) \\ &=\measuredangle DST - 90^\circ \\ &=\measuredangle HST =\measuredangle HAT =\measuredangle CDB =\measuredangle CNM, \\ \end{align*}so $M, O, N$ are collinear. Now let $P$ be the midpoint of $BD.$ We then have $\measuredangle MOB + \measuredangle BOP = 90^\circ = \measuredangle POD + \measuredangle DON,$ but since $OP$ perpendicularly bisects $BD$ we get $\measuredangle BOP = \measuredangle POD,$ so $\measuredangle MEB = \measuredangle MOB = \measuredangle DON = \measuredangle DEN,$ which means that $\measuredangle NEB = \measuredangle NEM + \measuredangle MEB = \measuredangle NEM + \measuredangle DEN = \measuredangle DEM.$ Finally, observe that \[\measuredangle HET = \measuredangle HAT = \measuredangle CAB = 90^\circ - \measuredangle ABD = \measuredangle TDO = \measuredangle TEO,\]so $E, H, O$ are collinear as well. However, $O$ is by definition on the perpendicular bisector of $TD,$ so $EO = EH$ bisects $\angle TED,$ and by symmetry $\angle BES$ as well. Thus, $E$ satisfies both conditions posed by the problem and we're done. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.82, xmax = 11.82, ymin = -7.33, ymax = 7.33; /* image dimensions */ /* draw figures */ draw(circle((-0.91,-0.7449097065462755), 5.223388791857912), linewidth(0.7)); draw(circle((-3.85,0), 2.23), linewidth(0.7)); draw(circle((1.32,-0.7449097065462751), 5.951746841970586), linewidth(0.7)); draw(circle((-2.7549771209025233,-1.8524097065462661), 4.222794665370196), linewidth(0.7)); draw(circle((-2.0993920068027223,0.7350451467268607), 3.7259237866484956), linewidth(0.7)); draw((-1.62,4.43)--(-1.62,-5.919819413092551), linewidth(0.7)); draw((-1.62,-5.919819413092551)--(4.26,0), linewidth(0.7)); draw((4.26,0)--(-1.62,4.43), linewidth(0.7)); draw((-1.62,4.43)--(-6.08,0), linewidth(0.7)); draw((-6.08,0)--(4.26,0), linewidth(0.7)); draw((-1.62,-5.919819413092551)--(-6.08,0), linewidth(0.7)); draw((-4.465088848010953,-2.143493808960632)--(1.32,2.215), linewidth(0.7)); draw((-3.834949552718485,2.229949211089039)--(1.32,-2.9599097065462754), linewidth(0.7)); draw((1.32,2.215)--(1.32,-2.9599097065462754), linewidth(0.7)); draw((1.32,-0.7449097065462751)--(-5.810954605207025,1.0618648860930264), linewidth(0.7)); draw((-1.62,-0.7449097065462755)--(1.32,-0.7449097065462751), linewidth(0.7)); /* dots and labels */ dot((-6.08,0),dotstyle); label("$A$", (-6.64,-0.59), NE * labelscalefactor); dot((4.26,0),dotstyle); label("$C$", (4.54,-0.43), NE * labelscalefactor); dot((-1.62,4.43),dotstyle); label("$B$", (-1.84,4.69), NE * labelscalefactor); dot((-1.62,0),linewidth(4pt) + dotstyle); label("$H$", (-1.6,0.39), NE * labelscalefactor); dot((1.32,2.215),linewidth(4pt) + dotstyle); label("$M$", (1.4,2.37), NE * labelscalefactor); dot((-1.62,-5.919819413092551),linewidth(4pt) + dotstyle); label("$D$", (-1.88,-6.61), NE * labelscalefactor); dot((1.32,-2.9599097065462754),linewidth(4pt) + dotstyle); label("$N$", (1.44,-3.35), NE * labelscalefactor); dot((-4.465088848010953,-2.143493808960632),linewidth(4pt) + dotstyle); label("$S$", (-5.02,-2.79), NE * labelscalefactor); dot((-3.834949552718485,2.229949211089039),linewidth(4pt) + dotstyle); label("$T$", (-4.22,2.49), NE * labelscalefactor); dot((1.32,-0.7449097065462751),linewidth(4pt) + dotstyle); label("$O$", (1.6,-0.97), NE * labelscalefactor); dot((-5.810954605207025,1.0618648860930264),linewidth(4pt) + dotstyle); label("$E$", (-6.38,1.11), NE * labelscalefactor); dot((-1.62,-0.7449097065462755),linewidth(4pt) + dotstyle); label("$P$", (-1.5,-1.27), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Remark: It turns out that $OH, DM, BN$ concur at the centroid of $\triangle BCD$ as well, which is the ISL 2011/G4 configuration.

Let $E = \odot(ABCD) \cup \odot(ATS).$ We'll show that E is the desired point. Claim. $MS \perp AD$ and $NT \perp AB$.

Claim. $H \in \odot(ATS)$. Furthermore, $BD$ is the tangent to the $\odot(ATS)$.

Claim. $EH$ bisects both angles $\angle BES$ and $\angle TED$.

Claim. $B$, $M$, $S$, $E$ are concyclic.

Claim. $B$, $D$, $S$, $T$ are concyclic.

Claim. $T$, $N$, $D$, $E$ are concyclic.

Claim. $\angle BEN = \angle MED$.

Thus, we are done by previous claims. $\blacksquare$

Nice problem! By angle chasing, $\angle ASH=\angle ATH=90^\circ$. Let $E$ be the intersection of the circumcircles of $ABCD$ and $ASHT$. We claim that $E$ is the desired point. Claim 1: The circumcircle of $\triangle BEH$ is tangent to $\overline{CB}$. Proof: We have \[\measuredangle DHE=\measuredangle HAE=\measuredangle CDE,\]so the circumcircle of $\triangle BEH$ is tangent to $\overline{CB}$. Thus, \[\measuredangle BEH=\measuredangle CBD=\measuredangle CAD=\measuredangle HES,\]so $\overline{EH}$ bisects $\angle BES$. Similarly, $\overline{EH}$ bisects $\angle TED$. Claim 2: $BESM$ and $DETN$ are cyclic. Proof: We have \[\measuredangle EBM=\measuredangle EAC=\measuredangle ESM,\]so $BESM$ is cyclic. Similarly, $DETN$ is cyclic. Claim 3: $BTSD$ is cyclic. Proof: There is a direct angle chasing proof, but the quickest way I've found is power of a point with \[AD \cdot AS=AH^2=AB \cdot AT.\] Thus, we have \[\measuredangle BEM=\measuredangle BSM=\measuredangle BSD-90^\circ=\measuredangle BTD-90^\circ=\measuredangle NTD=\measuredangle NED,\]so $\measuredangle BEN=\measuredangle MED$.

Same as @above, @2above... Claim: $ASHT$ and $STMN$ are cyclic. Proof. Notice $$\measuredangle HAT+\measuredangle THA=\measuredangle CDB+\measuredangle NHC=\measuredangle CDB+\measuredangle HCN=90$$as $NH=NC.$ Thus, $\measuredangle ATN=90$ similarly $\measuredangle HSA=90.$ Then, $$\measuredangle SMN=\measuredangle MHB=\measuredangle SHD=\measuredangle SAH=\measuredangle STH.$$$\blacksquare$ Let $E=(ABCD)\cap(ASHT).$ We claim $E$ is the desired point. Claim: $\overline{EH}$ bisects $\angle BES.$ Proof. Since $$\measuredangle EHS=\measuredangle EAS=\measuredangle EBD=\measuredangle EBH,$$$\overline{SM}$ is tangent to $(EHB)$ at $H.$ Hence, $$\measuredangle SEH=\measuredangle SAH=\measuredangle SHD=\measuredangle MHB=\measuredangle HEB.$$$\blacksquare$ Similarly, $\overline{EH}$ bisects $\angle TED.$ Claim: $DETN$ is cyclic. Proof. Note $\measuredangle NDE=\measuredangle CAE=\measuredangle ETH.$ $\blacksquare$ Similarly, $BESM$ is cyclic. Claim: $BDST$ is cyclic. Proof. We see $$\measuredangle DST=\measuredangle MST+90=\measuredangle MNT+90=\measuredangle 90-NHC+90=-\measuredangle HCN=\measuredangle DBA.$$$\blacksquare$ Then, $$\measuredangle DEN=\measuredangle DTN=\measuredangle DTB-90=\measuredangle DSB-90=\measuredangle MSB=\measuredangle MEB.$$$\square$

sorry I'm too lazy to turn all "\dang"s into "\measuredangles" and fix it so it looks good on aops

This problem is an actual tragedy. Let $F$ be the point on $(ABCD)$ such that $\overline{BD} \parallel \overline{CF}$, and let $E$ be the second intersection of $\overline{FH}$ with $(ABCD)$. I claim that this works. Clearly $\overline{MHS} \perp \overline{AD}$ and $\overline{NHT} \perp \overline{AB}$. By inversion at $H$ fixing $(ABCD)$, we find that $S$ and $T$ get sent to the reflections of $H$ over the $M$ and $N$ respectively. Call these points $S'$ and $T'$, so $BDT'S'$ is a rectangle. Now, $BHT'F$ is a parallelogram, so $\measuredangle HBF=\measuredangle FT'H$. Inverting back, this means $\measuredangle HED=\measuredangle TEH$, which is the desired bisection; the analogous case for $\angle BES$ is true as well. Observe now that $E$ is in fact the "ISL 2011/G4" point of $\triangle CBD$, hence $(EMN)$ is tangent to $(ABCD)$. Thus a homothety implies that the line between the second intersections of $\overline{EM}$ with $\overline{EN}$ is parallel to $\overline{BD}$, i.e. $\measuredangle BEM=\measuredangle NED$ as needed. $\blacksquare$

Invert at $H$ and refer to all points' images (other than $H$) by their original names. We claim that when $E$ is the antipode of $C$ in $(ABCD)$, both conditions hold. The first condition translates to $\angle EBH = \angle ESH$ and $\angle EDH = \angle ETH$. Note that $BTDS$ is a rectangle, with $E$ and $H$ symmetric about its center – therefore, by symmetry this condition is obviously satisfied. The second condition translates to \[\angle EBH + \angle ENH = \angle EDH + \angle EMH,\]assuming that $H$ lies inside $\triangle EMN$. We have $\triangle ECM \cong \triangle ECN$ since $CM = CN$ and $\angle MCE = \angle BCD = \angle ECN$. With this, we have $\angle ENC = \angle EMC$ and the second condition follows from a hefty angle chase.

First, Claim: $HT \perp AB$ and $HS \perp AD$. Proof. Note \[ \measuredangle THA + \measuredangle HAT = \measuredangle NHC + \measuredangle BAC = \measuredangle HCD + \measuredangle BDC = 90 \]the other is similar. $\blacksquare$ Thus $ATHS$ is cyclic. Let $(ATHS)$ intersect $(ABCD)$ at $E$. I claim that this is the desired point. Claim: $EH$ bisects $\angle BES$ and $\angle TED$. Proof. Chase \[ \measuredangle BEH = \measuredangle BEA + \measuredangle AEH = \measuredangle BDA - 90 = \measuredangle HAD = \measuredangle HAS = \measuredangle HES \]the other follows by similar logic. $\blacksquare$ Now, Claim: If we show that $(EMN)$ is tangent to $(ABCD)$ at $E$ then we're done. Proof. Assume that $(EMN)$ and $(ABCD)$ are tangent to $E$. Consider the homothety sending $(EMN)$ to $(ABCD)$ centered at $E$. Then it sends $M$ to a point $M'$ on $(ABCD)$ and $N$ to a point $N'$ on $(ABCD)$. But $M'N' \parallel MN \parallel BD$, and so $BM'N'D$ is an isosceles trapezoid. Thus \[ \measuredangle BEM = \measuredangle BEM' = \measuredangle N'ED = \measuredangle NED \]which implies $\angle BEN = \angle MED$. $\blacksquare$ There are two ways to finish now. The first is to observe that we can apply Shortlist 2011 G4 to $\triangle CBD$. The other is via inversion. Consider an inversion at $H$ with radius $\sqrt{HA \cdot HC}$. This inversion swaps $A$ and $C$, swaps $B$ and $D$, and swaps $E$ and the $A$-antipode with respect to $(ABCD)$, which we denote $A'$. Furthermore, it sends $M$ to the reflection of $H$ across $AD$, which we denote $X$, and it also sends $N$ to the reflection of $H$ across $AB$, which we denote $Y$. It suffices to prove $(A'XY)$ is tangent to $(ABCD)$. Note $AH = AX = AY$ and $XY \perp AA'$, since $XY$ and $BD$ are antiparallel with respect to $AB$ and $AD$. Then $AA'$ is the perpendicular bisector of $XY$, hence the center of $(A'XY)$ lies on $AA'$. This essentially finishes the problem, because then considering the homothety at $A'$ sending the center of $(A'XY)$ to the midpoint of $AA'$ shows that $(A'XY)$ and $(ABCD)$ are tangent. $\blacksquare$