It's a really cute problem

Let $P$ be the intersection of $AC$ and $BD$ and let $\angle BAC = \angle BCA = \alpha, \angle CAD = \angle CDA = \beta$. Through simple angle chasing we find that $\angle EAP = \angle ABP = 180^{\circ} - 2\beta - \alpha$ and $\angle EDP = \angle DCP = 180^{\circ} - 2\alpha - \beta$. We may now note that $EP$ is perpendicular to $BC$ if and only if $\angle EPA = 90^{\circ} - \alpha$, which happens if and only if $\angle AEP = 2\alpha + 2\beta - 90^{\circ}$, if and only if $EP$ bisects angle $AED$. We now prove that $P$ has the same distance to $EA$ and $ED$, which implies the result. Looking at triangles $EAP$ and $EDP$ we find that $d(P, EA) = AP\sin(\alpha + 2\beta)$ and $d(P, ED) = DP\sin(2\alpha + \beta)$. We now wish to prove that $$\frac{AP}{\sin(2\alpha + \beta)} = \frac{DP}{\sin(\alpha + 2\beta)}$$ However, this follows from the Law of Sines applied to triangles $APB, BPC$ and $CPD$, because $$\frac{AP}{\sin(2\alpha + \beta)} = \frac{BP}{\sin \alpha} = \frac{CP}{\sin \beta} = \frac{DP}{\sin(\alpha + 2\beta)}.$$

A bit hard for G1. Let $P=AB\cap CD,\ Q=BC\cap ED,\ R=BC\cap EA,\ T=AC\cap BD$. The angle conditions imply that $BDQP$ and $CARP$ is cyclic. Combining with $AB=BC=CD$ gives $PC=CQ$ and $PB=BR$. Since $AB=BC$, line $AT$ is parallel to angle bisector of $\angle PBC$ so if we let $I$ be the incenter of $\Delta PBC$, we get that $BICT$ is parallelogram. Thus the foot $K$ from $T$ to $BC$ is the contact point of $P$-excircle of $\Delta PBC$. Since $\angle EQR=\angle BPC =\angle ERQ$, we get that $\Delta EQR$ is isosceles. Furthermore, length chasing reveals that $$QK = QC + CK = PC + \frac{PB + BC - PC}{2} = \frac{QR}{2}$$hence $QK=KR\implies EK\perp QR$. But $TK\perp QR$ so we are done.

Extend $AB$ and $CD$ to meet at $R$, extend $AE$ and $BC$ to meet at $X$ and extend $ED$ and $BC$ to meet at $Y$. We get from the conditions, $\triangle AXB\cong CRB\cong CYD$. Suppose $T$ is the feet from $E$ on $BC$, $XT=YT$. Also suppose $Z=AC\cap BD$ and the feet from $Z$ on $BC$ is $T'$. Angle chasing gives $T'$ is the $P-$ excircle touchpoint with $BC$. Thus \[BT'=\frac{BC+PC-PB}{2}\]and \[BT=\frac{QR}{2}-BR=\frac{BC+PC+PB}{2}-BP=\frac{BC+PC-PB}{2}.\]Thus $T'\equiv T$.

Let $P = AC\cap BD$. Let $H$ be the orthocenter of triangle $BCP$. Claim 1: $D$ is the reflection of $B$ in $\overline{CH}$ and $A$ is the reflection of $C$ in $\overline{BH}$. Proof: $AB = BC$ and $AC \perp BH$. So, we are pretty much done. Claim 2: $H$ lies on the angle bisector of $\angle AED$. Proof: By the above, $\triangle ABH \cong \triangle CBH \cong \triangle CDH$. So, \[\angle EDH = \angle EDC - \angle CDH = \angle ABC - \angle ABH = \angle ABH = \angle CDH\]Thus, $H$ lies on the angle bisector $\angle CDE$. Similarly, for $\angle BAE$. Thus, by the property of angle bisectors, \[d(H, \overline{AE}) = d(H, \overline{AB}) = d(H, \overline{BC}) = d(H, \overline{CD}) = d(H, \overline{DE}).\]Therefore, $H$ lies on the angle bisector of $\angle AED$. Claim 3: $P$ lies on the angle bisector of $\angle E$. By Sine Rule, \[\frac{AP}{\sin\angle PDE} = \frac{AP}{\sin\angle ABP} = \frac{BP}{\sin \angle BAP} = \frac{BP}{\sin \angle BCP} = \frac{CP}{\sin\angle CBP} = \frac{CP}{\sin\angle CDP} = \frac{DP}{\sin\angle PCD} = \frac{DP}{\sin\angle PAE}\]\[\iff d(P, \overline{AB}) = AP\sin\angle PAE = DP\sin\angle PDE = d(P, \overline{ED})\] Main Proof: Since, $P, H, E$ are collinear and $PH\perp BC$, we are done. Any non-trig solutions of claim 3?

We present three solutions. [asy][asy] pair T = dir(270); pair S = dir(215); pair U = dir(10); pair R = S*T/U; pair V = T/S*U; pair A = 2*R*S/(R+S); pair B = 2*S*T/(S+T); pair C = 2*T*U/(T+U); pair D = 2*U*V/(U+V); pair E = 2*V*R/(V+R); draw(A--B--C--D--E--cycle, heavygreen); pair X = extension(D, E, A, B); pair Y = extension(D, C, A, E); draw(A--C, lightblue); draw(B--D, lightblue); pair P = extension(A, C, B, D); pair H = orthocenter(B, P, C); draw(B--Y, red); draw(C--X, red); draw(A--X--E--Y--D, dashed+heavygreen); draw(unitcircle, blue+dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$P$", P, dir(P)); dot("$H$", H, dir(180)); /* TSQ Source: T := dir 270 S := dir 215 U := dir 10 R := S*T/U V := T/S*U A = 2*R*S/(R+S) B = 2*S*T/(S+T) C = 2*T*U/(T+U) D = 2*U*V/(U+V) E = 2*V*R/(V+R) A--B--C--D--E--cycle 0.1 lightgreen / heavygreen X = extension D E A B Y = extension D C A E A--C lightblue B--D lightblue P = extension A C B D H = orthocenter B P C R180 B--Y red C--X red A--X--E--Y--D dashed heavygreen unitcircle blue dotted */ [/asy][/asy] First solution by Brianchon Draw the circle tangent to $AB$, $BC$, $CD$ inside the pentagon; then the angle conditions guarantee the incircle is tangent to $EA$ and $ED$ as well. Now apply Brianchon theorem. Second solution by Pappus Theorem (mine) Let the diagonals $AC$ and $BD$ meet at $P$. Let $H$ be the orthocenter of $\triangle BCP$. As $BH$ and $CH$ are perpendicular bisectors of $AC$ and $BD$, it follows that $BH$, $AE$, $CD$ are concurrent at $Y$, say. Similarly, $CH$, $DE$, and $AB$ are concurrent at a point $X$. By Pappus theorem on $\overline{XAB}$ and $\overline{YCD}$, it follows that $H$, $P$, $E$ are collinear, as desired. Third solution by trig (with Ishika Shah) Again let $P = \overline{AC} \cap \overline{BD}$. Let $T$ be the foot from $P$ to $\overline{BC}$. Let $\alpha = \angle BAC = \angle BCA$, and $\beta = \angle CBD = \angle BDC$. [asy][asy] pair T = dir(270); pair S = dir(215); pair U = dir(10); pair R = S*T/U; pair V = T/S*U; pair A = 2*R*S/(R+S); pair B = 2*S*T/(S+T); pair C = 2*T*U/(T+U); pair D = 2*U*V/(U+V); pair E = 2*V*R/(V+R); draw(A--B--C--D--E--cycle, heavygreen); draw(A--C, lightblue); draw(B--D, lightblue); pair P = extension(A, C, B, D); draw(unitcircle, blue+dotted); pair U = extension(A, E, B, C); pair V = extension(D, E, B, C); draw(A--U--B, dotted+blue); draw(D--V--C, dotted+blue); pair T = foot(P, B, C); draw(P--T, red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$P$", P, dir(95)); dot("$U$", U, dir(U)); dot("$V$", V, dir(V)); dot("$T$", T, dir(T)); /* TSQ Source: T := dir 270 S := dir 215 U := dir 10 R := S*T/U V := T/S*U A = 2*R*S/(R+S) B = 2*S*T/(S+T) C = 2*T*U/(T+U) D = 2*U*V/(U+V) E = 2*V*R/(V+R) A--B--C--D--E--cycle 0.1 lightgreen / heavygreen A--C lightblue B--D lightblue P = extension A C B D R95 unitcircle blue dotted U = extension A E B C V = extension D E B C A--U--B dotted blue D--V--C dotted blue T = foot P B C P--T red */ [/asy][/asy] Let $E_1$ be a point on $\overline{PT}$ satisfying $\angle E_1AB = \angle C = \pi - 2 \beta$. Similarly, let $E_2$ be a point on $\overline{PT}$ satisfying $\angle E_2DC = \angle B = \pi - 2 \alpha$. Let $U = \overline{AE_1} \cap \overline{BC}$ and $V = \overline{DE_2} \cap \overline{BC}$. By angle chasing we see that $\angle BUA = \angle DVC = 180^{\circ} - (2\alpha + 2\beta)$. We just need to show that $UT = TV$ now. Then, that will imply $E = E_1 = E_2$. WLOG, scale to let \[ AB = BC = CD = \sin(2\alpha+2\beta). \]Then, by sine law on $\triangle AUB$, we get $BU = \sin(2\beta)$. Meanwhile, by the law of sines on $\triangle BPC$ we have \[ BP = \frac{BC}{\sin \angle BPC} \cdot \sin \angle PCB = \frac{\sin(2\alpha+2\beta)}{\sin(\alpha+\beta)} \cdot \sin\alpha. \]Then, \[ BT = BP \cos \beta = \frac{\sin(2\alpha+2\beta) \sin \alpha \cos \beta}{\sin (\alpha+\beta)} \]and hence we have \begin{align*} TU &= TB + BU \\ &= \frac{\sin(2\alpha+2\beta) \sin \alpha \cos \beta} {\sin(\alpha+\beta)} + \sin(2\beta) \\ &= 2\cos(\alpha+\beta) \sin \alpha \cos \beta + \sin(2\beta) \\ &= 2\cos(\alpha+\beta) \sin \alpha \cos \beta + 2\sin\beta\cos\beta \\ &= 2[\cos\alpha\cos\beta-\sin\alpha\sin\beta] \sin \alpha \cos \beta + 2\sin\beta\cos\beta \\ &= 2[\cos^2\beta \cos\alpha\sin\alpha + (1-\sin^2\alpha)\sin\beta\cos\beta] \\ &= 2[\cos^2\beta \cos\alpha\sin\alpha + \cos^2\alpha \sin\beta\cos\beta]. \end{align*}This is symmetric in $\alpha$ and $\beta$, so \[ TV = 2[\cos^2\beta \cos\alpha\sin\alpha + \cos^2\alpha \sin\beta\cos\beta] \]as well. This finishes the proof.

Let $F=AE\cap BC,G=DE\cap BC,H=AC\cap BD,P=AB\cap CD$ and I be a excenter of $\Delta BCP$ opposite P. Angle chasing reveals that $\Delta EFG$ is isoscele.Because $AB=BC$,$\angle BAE=\angle BCD$ and $\angle ABF =\angle CBP$ so $\Delta ABF \cong \Delta BCP$.Similarly $\Delta ABF \cong \Delta CBP \cong \Delta CDG$.$IB$ bisects $\angle ABC$ so $IB$ is axis of reflection $\Delta ABF$ to $\Delta CBP$.We get $IP$ bisects $\angle BPC$ \rightarrow $IF$ bisects $\angle AFB$.In the same way,$IG$ bisects $\angle CGD$. $\therefore I$ is incenter of $\Delta EFG$.This mean $EI\perp BC$. Because $\Delta ABC ,\Delta BCD$ are isoscele so $IB\perp AC$ and $IC\perp BD$. $\therefore H$ is orthocenter of $\Delta BIC$ so $IH\perp BC$.So we get that $E,I,H$ are collinear and this line perpendicular to $BC$. $\square$

Let $P = AC \cap BD, X = ED \cap BC, Y = EA \cap CB$ Note that triangle $EXY$ is $E-isosceles$. Now, it would suffice to show that $P$ is equidistant from $YE$ and $XE$. Consider the triangles $ABP$ and $DCP$. Using the extended sine law, they have equal circumradii, and thus $DP sin ABP = AP sin PCD$ which means $DP sin PDE = AP sin PAE$. And we're done. $\blacksquare$

Let $X,Y$ be the intersections of lines $\overleftrightarrow{EA},\overleftrightarrow{ED}$ with $\overleftrightarrow{BC},$ respectively. Further let $K=\overline{AC}\cap \overline{BD},$ and let $T$ be the spiral center sending $\overline{CD}\to \overline{AB}.$ Lemma: $E,T$ lie on the perpendicular bisector of $\overline{XY}.$ Proof: The angle conditions as well as $CD=AB$ imply that $\triangle CDY\cong \triangle ABX.$ Thus $$\angle EYX=\angle DYC=\angle BXA=\angle EXY,$$implying the first part. For the second part, note that as $T$ is the spiral center that sends $\overline {CD}\to \overline {AB},$ it also takes the corresponding triangles $\triangle CDY\to \triangle ABX,$ whence $TX=TY$ follows. Now, notice that as $\overline{TB}$ is the perpendicular bisector of $\overline{AC}$ and $TC$ is the perpendicular bisector of $\overline{BD},$ $K$ is the orthocenter of $\triangle TBC\implies \overline{TK}\perp \overline{XY}.$ We may now conclude, since $E,T,K$ lie on the perpendicular bisector of $\overline{XY}\implies \overline{EK}\perp \overline{BC}.$

Suppose $CD$ intersects $AB$ at $P$ and $AE$ at $Q$, and let the foot of the perpendicular from $A$ be $R$. Now, let $BC=CD=DE=c$, $BP=DQ=a$, $PC=EQ=b$, $\angle PBC=\alpha$, and $\angle PCB=\beta$. We wish to show that $\frac{b+c-a}{2}\tan\frac{\alpha}{2}=\frac{c+a-b}{2}\tan\frac{\beta}{2}$, or $\frac{b+c-a}{2}\frac{\sin\alpha}{1+\cos\alpha}=\frac{c+a-b}{2}\frac{\sin\beta}{1+\cos\beta}$. Using law of sines as well as law of cosines, this is equivalent to $\frac{b+c-a}{2}\frac{2abc}{b^2+2bc+c^2-a^2}=\frac{a+c-b}{2}\frac{2abc}{a^2+c^2-2ac-b^2}\iff \frac{abc}{a+b+c}=\frac{abc}{a+b+c}$, which is true.

Here's another: Let $I$ be the intersection point of the angle bisector $\angle ABC$ and $\angle BCD$. Note that $B$ and $D$ are symmetric in $\overline{CI}$ and, $A$ and $C$ are symmetric in $\overline{AI}$. Claim 1: $I$ lies on the angle bisector of $\angle BAE$ and $\angle CDE$. Proof. \[2\angle CDI = 2\angle IBC = \angle ABC = \angle CDE.\]Similarly for $\angle BAE$. Claim 2: $I$ lies on the angle bisector of $\angle AED$. Proof. By the property of angle bisectors, \[d(I, \overline{AE}) = d(I, \overline{AB}) = d(I, \overline{BC}) = d(I, \overline{CD}) = d(I, \overline{DE}).\]Claim 3: $\overline{EI}\perp \overline{BC}$ Proof. Let $\overline{EI}\cap\overline{BC} = T$. \[\angle BTE = 360^{\circ} - \angle TBA - \angle BAE - \angle BEI = 360^{\circ} - \frac12 \sum_{\text{cyc }\in\{A,B,C,D,E\}} \angle ABC = 90^{\circ}.\] Now we can proceed from here in 3 ways. Proof 1: Since, $I$ is the concurrency point of the angle bisectors of the 5 angles and $IT\perp BC$, pentagon $ABCDE$ is circumscribed about $\odot(I, \overline{IT})$. Thus, by Brianchon's Theorem on the degenerate hexagon $ATBCDE$, $\overline{AC}$, $\overline{BD}$ and $\overline{ET}$ are concurrent and we are done. Let $P = \overline{AC}\cap \overline{BD}$. Proof 2: Since, $B$ and $D$ are reflections of each other in $\overline{CI}$, $\overline{BD}\perp\overline{CI}$. Similarly, $\overline{AC}\perp\overline{BI}$. Thus, $P$ is the orthocenter of $\triangle BCI$. This implies that $\overline{IP}\perp\overline{BC}$. However, we have $\overline{IE}\perp\overline{BC}$. Therefore, $\overline{EP}\perp\overline{BC}$ and we are done. Proof 3: By Law of Sines, \begin{align*} \frac{AP}{\sin\angle PDE} = \frac{AP}{\sin\angle ABP} = \frac{BP}{\sin \angle BAP} = \frac{BP}{\sin \angle BCP} &= \frac{CP}{\sin\angle CBP} = \frac{CP}{\sin\angle CDP} = \frac{DP}{\sin\angle PCD} = \frac{DP}{\sin\angle PAE}\\ \iff d(P, \overline{AB}) = AP\sin\angle PAE &= DP\sin\angle PDE = d(P, \overline{ED}) \end{align*}Thus, $P$ lies on the angle bisector of $\angle AED$, thus, $E, I, P$ are collinear. Therefore, $\overline{EP}\perp\overline{BC}$ and we are done.

[asy][asy] size(8cm); pair A = (1,1); pair B = (-1,1); pair C=B+2*dir(-140); pair D = C+2*dir(-20); pair X = A+dir(-120); pair Y = D+dir(20); pair E = extension(A,X,D,Y); draw(A--B--C--D--E--cycle, blue); pair P = extension(D,E,A,B); pair Q = extension(A,E,C,D); pair Z = extension(B,C,A,E); draw(B--Z^^A--Z, red); draw(A--P^^E--P^^E--Q^^D--Q, red); pair F = foot(E,B,C); pair H = A + dir(-150); pair O = extension(A,H,E,F); draw(circle(O, abs(O-F)), heavygreen); draw(B--D^^A--C, orange); draw(E--F, orange+dotted); dot("$A$", A, NE); dot("$B$", B, NW); dot("$C$", C, W); dot("$D$", D, SW); dot("$P$", P, NE); dot("$Q$", Q, S); dot("$Z$", Z, NE); dot("$F$", F, W); dot("$E$", E, SE); dot("$X$", extension(B,D,A,C), SE); [/asy][/asy] Let $P = AB \cap DE$ and $Q = AE \cap CD$. Since $CB = CD$ and $\angle ABC = \angle CDE$, we have $PB=PD$ so there exists a circle tangent to lines $\overleftrightarrow{AB}, \overleftrightarrow{BC}, \overleftrightarrow{CD}$, and $ \overleftrightarrow{DE}$. Similarly, we have $AQ=CQ$ so there exists a circle tangent to $\overleftrightarrow{AB}, \overleftrightarrow{BC}, \overleftrightarrow{CD}$, and $ \overleftrightarrow{AE}$. Hence, $ABCDE$ is a tangential pentagon with incircle $\omega$. Let $\angle BCD = \angle AEB = \alpha$, $\angle ABC = \angle CDE = \beta$, $Z = CB \cap AE$, and $F$ be the foot from $E$ to $BC$. Then \begin{align*} \angle FEA &= 90^\circ - \angle FZA\\ &= \angle ZBA + \angle ZAB - 90^\circ \\&= 270^\circ - \alpha - \beta \end{align*}but $\angle DEA = 540^\circ - (2\alpha + 2\beta)$ so $EF$ bisects $\angle DEA$. Thus, the center of $\omega$ lies on $EF$ so $\omega$ is tangent to $BC$ at $F$. Then by Brianchon's theorem on degenerate hexagon $ABFCDE$, lines $AC$, $BD$, and $EF$ are concurrent. $\square$

Let $AB \cap DC = P$, $EA \cap CB = Q$, $BA \cap DE = R$, $AE \cap CD = S$, $ED \cap BC = T$ Since $ \angle SAC = \angle ACS $ it's easy to note that $SA = SC$, thus, $BASC$ is a kite, and $SB \perp AC$ more important, $SB$ is the perpendicular bisector of $AC$. Analogously, $RC$ is the perpendicular bisector of $BD$. Let $Z = RC \cap SB$. Since $ \angle EAB = \angle BCD $, it's clear that $ \angle PCQ = \angle PAQ $ then $PQAC$ is cyclic and $ \angle AQC = \angle APC $ Since $ \angle ABC = \angle CDE $, it's clear that $ \angle CBP = \angle TDC $ then $PBDT$ is cyclic and $ \angle BTD = \angle BPD = APC $, Then $ \angle EQT = \angle AQC = \angle BTD = \angle QTE $, therefore $EQ=ET$, hence $E$ is part of the perpendicular bisector of $QT$ Since $BA=BC$ and $PQAC$ is cyclic, $ \angle QAC = \angle QPA = \angle CAP $ therefore $ PQ \parallel AC $ and since $ZB$ is perpendicular bisector of $AC$, then is perpendicular of $PQ$ too. Analogously, $ZC$ is perpendicular bisector of $PT$. Then $Z$ is circumcenter of $ \triangle PQT$, then $Z$ is on the perpendicular bisector of $QT$, therefore $ZE$ is part of the perpendicular bisector of $QT$, then $EZ \perp QT$, Now, notice what happens in $\triangle ZBC$, we have line $EZ \perp BC$, we have $BZ \perp AC$ and $CZ \perp BD$, then $BD$, $AC$ and $EZ$ concur at the orthocentre of $\triangle BCZ$, so we are done $Q.E.D.$

This is really nice! I'm going to steal parts of v_Enhance's diagram, since it has everything I need. [asy][asy] size(300); pair T = dir(270); pair S = dir(225); pair U = dir(-10); pair R = S*T/U; pair V = T/S*U; pair A = 2*R*S/(R+S); pair B = 2*S*T/(S+T); pair C = 2*T*U/(T+U); pair D = 2*U*V/(U+V); pair E = 2*V*R/(V+R); draw(A--B--C--D--E--cycle, heavygreen); draw(A--C, lightblue); draw(B--D, lightblue); pair P = extension(A, C, B, D); pair U = extension(A, E, B, C); pair V = extension(D, E, B, C); draw(A--U--B, dotted+blue); draw(D--V--C, dotted+blue); pair T = foot(P, B, C); draw(P--T, red); pair I1 = incenter(U,A,B), I2 = incenter(V,C,D); dot("$I_1$",I1,dir(90)); dot("$I_2$",I2,dir(90)); draw(I1--I2,orange+dotted); pair M = (B+D)/2, N = (A+C)/2, Z = (B+C)/2; draw(T--N--M--T,red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$P$", P, dir(95)); dot("$U$", U, dir(U)); dot("$V$", V, dir(V)); dot("$T$", T, dir(T)); dot("$M$",M,dir(90)); dot("$N$",N,dir(90)); draw(arc(Z,abs((B-C)/2),0,180), purple+dotted); [/asy][/asy] Let $U=EA\cap BC$, $V=ED\cap CB$, and $P=AC\cap BD$. Note that $\angle UAB = \angle DCV$ and $\angle UBA = \angle CDV$, so $\triangle UAB\cong\triangle VCD$. This in turn implies $UE=EV$. Now consider the spiral similarity sending $\triangle UAB$ to $\triangle VCD$. Let $T$, $M$, and $N$ be the midpoints of $\overline{UV}$, $\overline{BD}$, and $\overline{AC}$ respectively; then MGT dictates $\triangle TMN\sim\triangle UBA\sim\triangle VDC$. From $AB=BC=CD$ we have $BN\perp AC$ and $CM\perp BD$, so quadrilateral $BNMC$ is cyclic. Now a bit of angle chasing yields \[\angle NMP = \angle NCB = \tfrac12\angle ABU = \tfrac12\angle NMT\]and similarly $\angle PNM=\tfrac12\angle MNT$, so $P$ is the incenter of $\triangle TNM$. Now let $I_1$ and $I_2$ be the incenters of $\triangle UAB$ and $\triangle VCD$ respectively. Note that $UI_1 = VI_2$ and $\angle I_1UB = \angle I_2VU$, so $I_1I_2VU$ is an isosceles trapezoid. Now the Gliding Principle says that $P$ is the midpoint of $\overline{I_1I_2}$; combined with $T$ being a midpoint means that $PT$ is perpendicular to $UV$, and we're done!

Denote $P \equiv AC \cap BD, X \equiv BC \cap AE, Y \equiv BC \cap DE.$ Let $\ell$ be the perpendicular bisector of $\overline{XY}.$ We show that $PE \equiv \ell.$ The given conditions imply that $\triangle ABX \cong \triangle CDY.$ In particular, $\angle AXB = \angle CYD$, so $E \in \ell.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = 979.9461794369739, xmax = 1060.7064107876533, ymin = 1007.3646180586487, ymax = 1048.2285507881656; /* image dimensions */ pen evevff = rgb(0.8980392156862745,0.8980392156862745,1.); pen fsfsfs = rgb(0.9490196078431372,0.9490196078431372,0.9490196078431372); pen evefev = rgb(0.8980392156862745,0.9372549019607843,0.8980392156862745); pen qqwuqq = rgb(0.,0.39215686274509803,0.); filldraw((997.753378336808,1026.5161562605842)--(998.7739042990934,1015.6150834816251)--(990.9343767825453,1015.6546561791224)--cycle, evevff, linewidth(0.8) + blue); filldraw((1009.7225026821317,1015.5598166851355)--(1018.4457834116164,1022.1764044980849)--(1022.5469627287129,1015.4950808350579)--cycle, evevff, linewidth(0.8) + blue); filldraw((997.753378336808,1026.5161562605842)--(998.7739042990934,1015.6150834816251)--(1009.7225026821317,1015.5598166851355)--(1018.4457834116164,1022.1764044980849)--(1006.8691939691704,1041.036087286514)--cycle, fsfsfs, linewidth(0.8) + gray); filldraw((1008.8596390545646,1018.1468184382297)--(1009.6085646061937,1018.3966136374357)--(1009.3587694069878,1019.145539189065)--(1008.6098438553586,1018.8957439898589)--cycle, evefev, linewidth(0.8) + qqwuqq); filldraw((1004.271009043682,1021.6203309252821)--(1003.6886645912641,1022.1533994594946)--(1003.1555960570515,1021.5710550070768)--(1003.7379405094695,1021.0379864728642)--cycle, evefev, linewidth(0.8) + qqwuqq); 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((997.753378336808,1026.5161562605842)--(998.7739042990934,1015.6150834816251), linewidth(0.8) + blue); draw((998.7739042990934,1015.6150834816251)--(990.9343767825453,1015.6546561791224), linewidth(0.8) + blue); draw((990.9343767825453,1015.6546561791224)--(997.753378336808,1026.5161562605842), linewidth(0.8) + blue); draw((1009.7225026821317,1015.5598166851355)--(1018.4457834116164,1022.1764044980849), linewidth(0.8) + blue); draw((1018.4457834116164,1022.1764044980849)--(1022.5469627287129,1015.4950808350579), linewidth(0.8) + blue); draw((1022.5469627287129,1015.4950808350579)--(1009.7225026821317,1015.5598166851355), linewidth(0.8) + blue); draw((997.753378336808,1026.5161562605842)--(998.7739042990934,1015.6150834816251), linewidth(0.8) + gray); draw((998.7739042990934,1015.6150834816251)--(1009.7225026821317,1015.5598166851355), linewidth(0.8) + gray); draw((1009.7225026821317,1015.5598166851355)--(1018.4457834116164,1022.1764044980849), linewidth(0.8) + gray); draw((1018.4457834116164,1022.1764044980849)--(1006.8691939691704,1041.036087286514), linewidth(0.8) + gray); draw((1006.8691939691704,1041.036087286514)--(997.753378336808,1026.5161562605842), linewidth(0.8) + gray); draw((998.7739042990934,1015.6150834816251)--(1006.785049862698,1024.366765258095), linewidth(0.8) + linetype("2 2")); draw((1006.785049862698,1024.366765258095)--(1009.7225026821317,1015.5598166851355), linewidth(0.8) + linetype("2 2")); draw((998.7739042990934,1015.6150834816251)--(1018.4457834116164,1022.1764044980849), linewidth(0.8)); draw((1009.7225026821317,1015.5598166851355)--(997.753378336808,1026.5161562605842), linewidth(0.8)); draw((xmin, 198.10445110739707*xmin-198424.2329209251)--(xmax, 198.10445110739707*xmax-198424.2329209251), linewidth(0.8) + red); /* line */ /* dots and labels */ dot((998.7739042990934,1015.6150834816251),linewidth(3.pt) + dotstyle); label("$B$", (998.5917443709326,1014.3241402994681), S * labelscalefactor); dot((1009.7225026821317,1015.5598166851355),linewidth(3.pt) + dotstyle); label("$C$", (1009.3101529557233,1014.398573692418), S * labelscalefactor); dot((990.9343767825453,1015.6546561791224),linewidth(3.pt) + dotstyle); label("$X$", (990.0691208781651,1014.5474404783179), S * labelscalefactor); dot((1022.5469627287129,1015.4950808350579),linewidth(3.pt) + dotstyle); label("$Y$", (1022.5592969008117,1014.2497069065181), S * labelscalefactor); dot((997.753378336808,1026.5161562605842),linewidth(3.pt) + dotstyle); label("$A$", (997.028643118984,1027.0150337974328), N * labelscalefactor); dot((1018.4457834116164,1022.1764044980849),linewidth(3.pt) + dotstyle); label("$D$", (1018.5771103779902,1022.4001634345366), NE * labelscalefactor); dot((1006.8691939691704,1041.036087286514),linewidth(3.pt) + dotstyle); label("$E$", (1007.0399344707503,1041.7156289050458), NE * labelscalefactor); dot((1006.785049862698,1024.366765258095),linewidth(3.pt) + dotstyle); label("$I$", (1006.9282843813253,1024.5959485265598), NE * labelscalefactor); dot((1006.7543089439189,1018.2768524166869),linewidth(3.pt) + dotstyle); label("$P$", (1006.8910676848503,1018.4924103046648), NE * labelscalefactor); label("$\ell$", (1007.1888012566501,1035.3353500727665), NE * labelscalefactor,red); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $I$ be the center of rotation sending $\triangle ABX \mapsto \triangle CDY.$ Clearly $I \in \ell.$ From $IA = IC$, we infer that $IB$ is the perpendicular bisector of $\overline{AC}.$ Similarly, $IC$ is the perpendicular bisector of $\overline{BD}.$ Therefore, $P$ is the orthocenter of $\triangle IBC$, so $IP \perp BC.$ Thus, $P \in \ell.$

In @Vfire diagram: By Pitot’s theorem on the two kites $ABCQ$ and $BCDP$ gives that the pentagon has an inscribed circle. Brianchon then finishes the problem.

USJL wrote: It's a really cute problem

How can you prove $I$ to be on the angle bisector of $\angle E$ too and please explain the rest solution.

Durjoy1729 wrote: How can you prove $I$ to be on the angle bisector of $\angle E$ too and please explain the rest solution. Vrangr wrote: Claim 2: $I$ lies on the angle bisector of $\angle AED$. Proof. By the property of angle bisectors, \[d(I, \overline{AE}) = d(I, \overline{AB}) = d(I, \overline{BC}) = d(I, \overline{CD}) = d(I, \overline{DE}).\]

First, we claim that $ABCDE$ has an inscribed circle. To prove this, it suffices to show that and three of the interior angle bisectors concur. Luckily \[ \left( 180^{\circ} - \frac{1}{2} \angle A - \frac{1}{2} \angle B\right) + \left( 180^{\circ} - \frac{1}{2} \angle B - \frac{1}{2} \angle C\right) = 360^{\circ}-\left( \angle A / 2 + \angle B + \angle C / 2 \right), \]so the $A$, $B$, and $C$-bisectors concur. By Brianchon applied to $ABCDE$, the conclusion follows.

Since the internal angle bisectors of $\angle A$, $\angle B$, $\angle C$, and $\angle D$ clearly concur at the center of the circle tangent to sides $AB$, $BC$, $CD$, it is also tangent to $DE$ and $EA$. Let $M$ be the midpoint of $BC$. Then Brianchon's on $ABMCDE$ gives $AC$, $BD$ and $EM$ concurring.

Let $\angle A = 180-2x$, $\angle B = 180-2y$, $AB=BC=CD=1$, and suppose $F = EA \cap BC$ and $G = ED \cap BC$. We find that $\triangle FAB \sim \triangle GCD$ and that $\triangle EFG$ is isosceles. If we define $K = AC \cap BD$ and $L$ as the projection of $K$ onto $FG$, it suffices to show $FL = GL$, or \[FB+BL = \frac{\sin 2x}{\sin (2x+2y)} + \frac{\cos x \sin y}{\sin (x+y)} = \frac{\sin 2y}{\sin (2x+2y)} + \frac{\cos y \sin x}{\sin (x+y)} = GC+CL. \quad \blacksquare\]

let $AA'$ be a line parallel through $BD$ passing through $A$, and let $DD'$ be a line parallel through $AC$ passing through $D$ pretty easily, $ABD'=CBD=AD'B=ADB$, and $DCA'=BCA=DA'C=BAC$ we get that $ACA'$ and $DBD'$ are similar, and we can let $BD'$ intersect $AC$ at $F$, and let $BD$ intersect $AC'$ at $G$ since $FBG=FCG$, $BFGC$ is cyclic. $BGA'=CGD=180-GCD-GDC$, so $FGA'=180-2GCD-GDC$ and by previous similarities, $FGA'=BCA'=180-2GCD-GDC$ and $FG$ is parallel to $BC$, but since $BFGC$ is cyclic $BF=GC$ this implies $180-2GCD-GDC=180-GCD-2GDC$, and that $GCD=GDC$, so $ABC=BCD=CDE=EAB$, so then the conclusion is obvious since the perpendicular line from $E$ to $BC$ is a line of symmetry

Finally got around to solving this! Let $I$ be the intersection of the internal angle bisectors of $\angle ABC$ and $\angle BCD,$ and let $X = AC \cap BD.$ Then the length conditions translate into $B$ and $D$ being reflections across $IC,$ so $BD \perp IC.$ Similarly, $AC \perp IB,$ so $X$ is the orthocenter of $\triangle BIC.$ Thus $IX \perp BC.$ Moreover, $\angle BCI = \angle BAI$ because of the reflections. Multiplying this by $2$ gives $$2 \angle BAI = 2 \angle BCI = \angle BCD.$$But $\angle BCD = \angle EAB$ by the angle condition, so $\angle BAE = 2 \angle BAI.$ Thus $AI$ bisects $\angle BAE$ and similarly $DI$ bisects $\angle CDE.$ Therefore, there exists a circle centered at $I$ tangent to all sides of the pentagon $ABCDE.$ Thus if $T$ is the tangency point of this circle with $BC,$ a.k.a. the foot of the altitude from $I$ to $BC,$ Brianchon on $ABTCDE$ implies $E,X,T$ are collinear. However, since $X$ is the orthocenter of $\triangle BIC,$ we get $I,X,T$ are collinear, so $\overline{EX} = \overline{IT} \perp \overline{BC}$ as desired.

Remark: The main idea of this solution is that the given conditions look a lot nicer once $P$ is defined (from which $H$ naturally follows), and then adding points $A'$ and $D'$ makes the problem completely determined by five congruent triangles. Here's the restatement: Problem: Suppose five congruent triangles have the similarities \[ PD'A \overset{-}\sim PBA \overset{-}\sim PBC \overset{-}\sim PDC \overset{-}\sim PDA'. \]Then $AD' \cap A'D$ lies on the perpendicular from $P$ to $BC$.

let the perpendicular bisectors of $AC$ and $BD$ intersect at $Q$ and let $AC$ and $BD$ intersect at $P$ we easily prove that $APQY$ and $DPQC$ are cyclic let $BQ$ intersect $AC$ at $R$, let $CQ$ intersect $BD$ at $S$, and let $PQ$ intersect $BC$ at $T$ by radical axes and stuff, we get that $BRPT$ and $CSRT$ are cyclic, and $PQ$ is perpendicular to $BC$ (not done, later gonna prove $E$ lies on $PQ$ but I'm lazy rn)

Let $EA, ED$ meet $BC$ at $M,N$. Let $K$ be $AC \cap BD$, let $F$ be the foot from $K$ to $BC$, Clearly, $EMN$ is isosceles, so it suffices to prove $FM = FN$. Take $BK = BC \frac{\sin x}{\sin x + y}, BF = BC \frac{\sin x \cos y}{\sin x + y}$. Now $BM = AB \frac{\sin 2y}{\sin 2x + 2y} = BC \frac {\sin y \cos y}{\sin x + y \cos x + y}$. Now $$FM = BM + BF = \frac{BC \cos y}{\sin x + y} (\sin x +\frac{\sin y}{\cos x + y}) = \frac{BC \cos y}{\sin x + y} (\frac{\sin x \cos x \cos y - \sin x \sin x \sin y + \sin y}{\cos x + y}) =\frac{BC \cos y}{\sin x + y} (\frac{\sin x \cos x \cos y - (1 - \cos^2 x) \sin y + \sin y}{\cos x + y}) = \frac{BC \cos y}{\sin x + y} (\frac{\sin x \cos x \cos y + \cos x \cos x \sin y}{\cos x + y}) = \frac{BC \cos y}{\sin x + y} (\frac{\cos x \sin x + y}{\cos x + y}) = BC \frac{\cos x \cos y}{\cos x + y} $$, which is symmetric in $x,y$, so we are done.

Let $AC \cap BD = F$. Let $EF \cap BC = G$. We need to prove that $\angle EGC = 90^{\circ}$. Let $\angle EAB = \angle BCD = \alpha$ and $\angle ABC = \angle EDC = \beta$. Also $\angle AED = 540 - 2\alpha - 2\beta$ $\Rightarrow$ if EG is the angle bisector of $\angle AED$ it follows that $\angle GED = 270 - \alpha - \beta$ $\Rightarrow$ $\angle EGC = 360 - (270 - \alpha - \beta) - \alpha - \beta = 90^{\circ}$ $\Rightarrow$ we need to prove that EG is the angle bisector of $\angle AED$ and we would be ready. By law of sines on $\triangle FED$ we get that $\frac{FD}{\sin \angle FED} = \frac{FE}{\sin \angle FDE}$ $\Rightarrow$ $FE \cdot \sin \angle FED = FD \cdot \sin \angle FDE$. By law of sines on $\triangle FAE$ we get that $\frac{FE}{\sin \angle FAE} = \frac{AF}{\sin \angle AEF}$ $\Rightarrow$ $FE \cdot \sin \angle AEF = AF \cdot \sin \angle FAE$ $\Rightarrow$ to show that $\angle AEF = \angle FED$ it is enough to show that $FD \cdot \sin \angle FDE = AF \cdot \sin \angle FAE$. Now by law of sines starting from $\triangle FBA$ we get $\frac{FA}{\sin \angle FDE} = \frac{FA}{\sin \angle FBA} = \frac{AB}{\sin \angle AFB} = \frac{DC}{\sin \angle DFC} = \frac{FD}{\sin \angle FCD} = \frac{FD}{\sin\angle FAE}$ $\Rightarrow$ $\frac{FA}{\sin \angle FDE} = \frac{FD}{\sin\angle FAE}$ $\Rightarrow$ $FD \cdot \sin \angle FDE = FA \cdot \sin \angle FAE$ $\Rightarrow$ EF is the angle bisector of $\angle AED$ $\Rightarrow$ $\angle EGC = 90^{\circ}$ and we are ready.

Let $P = \overline{AE} \cap \overline{CD}$ and $Q= \overline{AB}\cap \overline{DE}$. Now, let $X = \overline{CQ} \cap \overline{BP}$. The entirety of the solution is the following key claim. Claim : Point $X$ is the incenter of tangential pentagon $ABCDE$. Proof : Since $X$ is the intersection of the diagonals of kites $PABC$ and $QBCD$, it is clear that it is the incenter of both these kites. Thus, $X$ lies on the $\angle B-$bisector and $\angle D-$bisector of $ABCDE$. Further, due to symmetry in kite $QBCD$, $BX=DX$. Thus in $\triangle BCX$ and $\triangle CDX$ we have $BX=XD$ , $BC = CD$ and $2\measuredangle XBC = \measuredangle ABC = \measuredangle CDE = 2\measuredangle CDX$ so $\measuredangle XBC = \measuredangle CDX$. This implies that $\triangle XBC \cong \triangle XDC$ and thus $X$ lies on the $\angle C-$bisector of $ABCDE$ as well. A similar argument shows that $X$ lies on the $\angle A-$bisector of $ABCDE$, which allows us to conclude that pentagon $ABCDE$ is tangential with $X$ being its incenter. Now, this implies that the $\angle E-$bisector must also pass thru $X$. Let $M$ be the foot of the altitude from $E$ to $\overline{BC}$. Note that since quadrilaterals $ABME$ and $CDEM$ have 3 pairs of equal angles, the other pair must also be equal, and thus $\measuredangle MEA = \measuredangle DEM$. So, $M$ is the tangency point of the incircle of $ABCDE$ with side $BC$. Now, by Brianchon's theorem on $ABMCDE$, it follows that diagonals $\overline{AC}$ and $\overline{BD}$ are concurrent with the perpendicular line from $E$ to $\overline{BC}$.

Faster triglen sol exploiting symmetry.

Let $AE \cap BC=X,AB \cap CD=Y,BC \cap DE=Z$ Let $\angle EAB=\angle BCD= \alpha, \angle ABC=\angle ECD=\beta$ Observe that $$\angle AXB=\angle BYC=\angle CZD=\alpha +\beta-180$$ALos noticing that $AB=BC=CD=a$, we get the fact that $AXYC$ and $BYZD$ are isosceles trapeziums (is it trapezia? idk) which also implies that $BX=BY=ZD=b $ and $AX=CY=CZ=c$. Also, $\angle FXZ=\angle FZX$ implies that $FX=FZ$. Let $\frac{a+b+c}{2}=s$. Let $Y$-excircle and $Y$-excentre of $\triangle YCB$ be $\omega$ and $O$ respectively. Let $\omega$ touch $BC, AB, CD$ in $P,Q,R$. Note that $P$ is midpoint of $XZ$, hence $FP \perp XZ \implies F-O-P$. Note that $BP=s-b=BQ$ and $CP=s-c=CR$, therefore $\omega $ is the $X$-excircle of $\triangle AXB$ and the $Z$-excircle of $\triangle ZCD$. Hence, $\omega$ is inscribed in $ABCDE$. Let $AC \cap OP=M$. let $I$ be the incentre of $\triangle YBC$ and $ K$ be the foot of the altitude from $I$ on $BC$. Note that $\angle PCM= 90-\frac{\beta}{2}=\angle IBK$, $\angle IKB=90=\angle MPC$ and $BK=CP$ $\implies \triangle IBK\cong \triangle MPC$, so $IB=MC$. Hence $BICM$ is a parallelogram and $\angle CBM=\angle IBC=90-\frac{\alpha}{2}=\angle DBC$. Hence, $AC\cap BD=M \in OP \in FP \perp BC$, and we are done. $\blacksquare$

Let $P = \overline{AE} \cap \overline{BC}$, $Q = \overline{DE} \cap \overline{BC}$, and $H$ be the foot of the perpendicular from $E$ to $\overline{BC}$. If we let $X_1 = \overline{AC} \cap \overline{EH}$ and $X_2 = \overline{BD} \cap \overline{EH}$, we find that $HX_1 = \tfrac{CH}{\tan(B/2)}$ and $HX_2 = \tfrac{BH}{\tan(C/2)}$. It then suffices to show that $HX_1 = HX_2$, or $\tfrac{BH}{CH} = \tfrac{\tan(C/2)}{\tan(B/2)}$. WLOG assume $AB = BC = CD = 1$. From the law of sines we have $PB = \tfrac{\sin C}{-\sin(B + C)}$ and $CQ = \tfrac{\sin B}{-\sin(B + C)}$. Since $\angle P = \angle Q = B + C - 180^{\circ}$, we have $PB + BH = CQ + CH$, so $$BH - \frac{\sin C}{\sin(B + C)} = CH - \frac{\sin B}{\sin(B + C)}.$$Combining the above equation with $BH + CH = 1$, we find that $BH = \tfrac{1}{2} \cdot \tfrac{\sin(B + C) + \sin C - \sin B}{\sin(B + C)}$ and $CH = \tfrac{1}{2} \cdot \tfrac{\sin(B + C) - \sin C + \sin B}{\sin(B + C)}$. Thus, $$\frac{BH}{CH} = \frac{\sin(B + C) + \sin C - \sin B}{\sin(B + C) - \sin C + \sin B}.$$By using sum to product identities, this equals \begin{align*} \frac{\sin(B + C) + \sin C - \sin B}{\sin(B + C) - \sin C + \sin B} &= \frac{2 \sin(\tfrac{B + C}{2})\cos(\tfrac{B + C}{2}) + 2 \sin(\tfrac{C - B}{2})\cos(\tfrac{B + C}{2})}{2 \sin(\tfrac{B + C}{2})\cos(\tfrac{B + C}{2}) + 2 \sin(\tfrac{B - C}{2})\cos(\tfrac{B + C}{2})} \\ &= \frac{\sin(\tfrac{B + C}{2}) + \sin(\tfrac{C - B}{2})}{\sin(\tfrac{B + C}{2}) + \sin(\tfrac{B - C}{2})} \\ &= \frac{2\sin(\tfrac{C}{2})\cos(\tfrac{B}{2})}{2\sin(\tfrac{B}{2})\cos(\tfrac{C}{2})} \\ &= \frac{\tan(\tfrac{C}{2})}{\tan(\tfrac{B}{2})}, \end{align*}so we are done.

Set $X= AE\cap BC$ and $Y=DE\cap BC$. Then $\triangle EXY$ is isosceles with $\triangle ABX\cong\triangle CDY$. In particular, let us set $AX=a$, $BX=b$, $AB=c$, $s=\frac{a+b+c}2$, $\alpha=\frac{\angle XBA}2$, and $\beta=\frac{\angle XAB}2$. Let $M$ be the midpoint of $XY$ and let $P=AC\cap BD$. It suffices to show that $P$ lies on $EM$. Considering $\triangle BCP$, this reduces to showing that $\frac{\tan\angle ACB}{\tan\angle DBC}=\frac{BM}{CM}$ or equivalently that $\frac{\tan\alpha}{\tan\beta}=\frac{s-b}{s-a}$. But in fact this identity always holds true for any triangle with angles $2\alpha$, $2\beta$, and $180^\circ-2(\alpha+\beta)$ and sides $a$, $b$, and $c$, as one can verify computationally (use Law of Sines to obtain the ratios $a/c$ and $b/c$ to write $\frac{s-b}{s-a}$ in terms of $\alpha$ and $\beta$, and then apply sum-to-product). $\blacksquare$ edit: just look at @above