Posted before, even quite recently (few months ago only!), if I am not wrong. Best regards, sunken rock

This problem is missing from the resource page.

It has now been corrected For reference, also posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=296040 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=12098

See and here.

Another way is possible also. You can use just a bit of complex numbers. Let $P$ be the origin, and $E$ be at $e$, $F$ at $f$. We have that for some $k$, $A=e+ike$ and $D=f-ikf$ because of the perpendicularity and $PFD$ is similar to $PEA$. Let the midpoint of $AD$ be $M$, which is half of $e+f+ike-ikf$. Now, $ME$ is half $-e+f+ike-ikf$ and $MF$ is half $e-f+ike-ikf$. These are $(e-f)(-1+ik)$ and $(e-f)(1+ik)$ respectively, and they have the same magnitude because $-1+ik$ and $1+ik$ have the same magnitude. Therefore, $ME=MF$, and the midpoint of $AD$ is on the perpendicular bisector of $EF$. Similar holds for the midpoint of $BC$ so we are done.

jgnr wrote: This problem is missing from the resource page.

Dayummm this is a nice solution. Although, what is the motivation for constructing the midpoints of S and R besides the right angles ad PF and PE?

Does anyone have a proof using spiral similarity?

bump $\text{ }$

^^^Here is a solution that is absolutely effortless. So, note that the line adjoining the midpoints is antiparallel to the line $XY$ where $X, Y = AC \cap BD, AB \cap CD$. But clearly, $XY$ is anti-parallel to the line perpendicular to $EF$, so the lines adjoining the midpoints must be perpendicular to $EF$ so done.

@mathisfun7 yes I found a spiral similarity solution last night! Let M be the midpoint of AD, N of BC, R of DP, and S of CP. I will prove that F is the center of the spiral smiilarity bringing quad. FRPS to FMEN. Since clearly RS||CD perpendicular to FP then this will solve the proof. Denote by X <BAC=<BDC (According to my diagram, yours might be flipped) <FRM=<FRD+<DRM=2(90-X)+<DPA=2(90-X)+<DPA, but <FPE=<FPE+<EPA+<EPA=2(90-X)+<DPA and you can say the same thing for <FSN. Now, we have FR/RM=(DP/2)/(AP/2)=DP/AP, FP/PE=(DPsinx)/(APsinx)=DP/AP and the same for FS/SN. So then FRM is similar to FPE is similar to FSN so F is the center of that spiral similarity, thus meaning that MN is perpendicular to FE.

Dear Mathlinkers, just a remark... EF is parallel to the Steiner's line of ABCD... Sincerely Jean-Louis

We proceed with complex numbers. Denote $P \equiv AC \cap BD$, and let $M, N$ be the midpoints of $\overline{AD}, \overline{BC}$, respectively. Then $2m = a + d, \; 2n = b + c$, and from the projection formula, we find $2e = a + b + p - ab\overline{p}, \; 2f = c + d + p - cd\overline{p}.$ To prove that $EF \perp MN$, we need only show that $\tfrac{e - f}{m - n} \in i\mathbb{R}.$ To see this, let \[z = \frac{e - f}{m - n} = \frac{a + b - c - d + (cd - ab)\overline{p}}{a + d - b - c}.\] From the intersection of chords formula, we have \[p = \frac{ac(b + d) - bd(a + c)}{ac - bd} \implies \overline{p} = \frac{\frac{1}{ac}\left(\frac{1}{b} + \frac{1}{d}\right) - \frac{1}{bd}\left(\frac{1}{a} + \frac{1}{c}\right)}{\frac{1}{ac} - \frac{1}{bd}} = \frac{b + d - a - c}{bd - ac}.\] Hence, \[z = \frac{a + b - c - d + (cd - ab)\left(\frac{b + d - a - c}{bd - ac}\right)}{a + d - b - c} = \frac{-a^2c + b^2d + ac^2 - bd^2 - ab^2 + cd^2 + a^2b - c^2d}{(bd - ac)(a + d - b - c)},\] where we have multiplied through by $bd - ac$, and expanded the numerator. This factors as \[z = \frac{(a + d - b - c)(ab + cd - ac - bd)}{(bd - ac)(a + d - b - c)} = \frac{ab + cd - ac - bd}{bd - ac}.\] It remains to show that $\overline{z} = -z$, which is just a straightforward computation..

Denote $P \equiv AC \cap BD$ and $X \equiv AB \cap CD.$ Let $Q$ be the reflection of $P$ in the midpoint of $\overline{BC}.$ We claim that $\triangle XBQ \sim \triangle XDP.$ Indeed, note that $BPCQ$ is a parallelogram because its diagonals bisect one another. Hence, \[\measuredangle XBQ = \measuredangle XAC = \measuredangle BAC = \measuredangle BDC = \measuredangle PDX,\] where the angles are directed. Now, let $A'$ be the reflection of $A$ in $E$, and note that $X, D, P, A'$ are concyclic because \[\measuredangle XDP = \measuredangle CDB = \measuredangle CAB = \measuredangle PAA' = \measuredangle XA'P.\] It follows that $\triangle BA'P \sim \triangle BDX$, and therefore $XD : XB = PA' : PB = PA : PB.$ Because $\triangle PAB \sim \triangle PDC$, we obtain $XD : XB = PD : PC = PD : BQ.$ Thus by side-angle-side similarity, we have $\triangle XBQ \sim \triangle XDP$ as desired. It follows that $XP$ and $XQ$ are isogonal w.r.t. $\angle BXC.$ Therefore, it is well-known that $XQ \perp EF.$ Then note that a homothety with center $P$ and ratio $\tfrac{1}{2}$ sends $XQ$ to the Newton-Gauss line of quadrilateral $XAPD.$ The desired result follows. $\square$

Let the midpoint of $BC \equiv M$ and the midpoint of $AD \equiv N$. Let $BD \cap AC \equiv P$. Denote the reflections of $P$ about $E$ and $F$ be $E'$ and $F'$ respectively. It is clear that $\triangle PCD \sim PBA$, so by definition of reflection $F'DPC$ and $PAE'B$ are kites similar to one another. Then since $M$ and $N$ are midpoints of $BC$ and $AD$, and since $E$ and $F$ are midpoints of $PE'$ and $F'P$, we have that $NFME$ is a kite similar to $F'DPC$ and $PAE'B$ by MGT, so $EF \perp MN$ as desired.

Let $G=AC\cap BD, H\in AB: \overline{HE}=\overline{EB}, I\in CD: \overline{IF}=\overline{FC}, M$ is the midpoint of $BC$ We have $\Delta GHC=\Delta GBI\Rightarrow ME=MF\Rightarrow Q.E.D$

The reflections of $A, D$ about $F, E$ are $S, T$ respectively. We have $\Delta APT=\Delta SPD\rightarrow SD=AT$. Let $M, N$ are midpoint of $AD, BC$ respecttively so $MF=DS/2=AT/2=ME$ Analogously,$NE=NF\rightarrow MN$ is perpencular bisector of EF

Let the diagonals meet at $P$ and $H_1,H_2$ be the orthocenters of $PDC,PAB$ it is not difficult to show $H_1H_2\parallel EF$ (just prove $\frac{PH_1}{PF}=\frac{PH_2}{PE}$) so $EF$ is parallel to the Steiner line which is perpendicular to the Gauss line which passes throw the midpoints of $AD,BC$

Let $AB\cap CD = P$, $AC\cap BD = K$ and $X, Y$ be the midpoints of $AD, BC$. Since $XY$ is Newton-Gauss line of $ABDC$, it pass through midpoint of $PK$, which is the center of $\odot(PKEF)$. Moreover, since $\triangle AKB\cup E\sim\triangle DKC\cup F$, we get $AE : EB = DF : FC$. Thus by E.R.I.Q. lemma, $XY$ also pass through midpoint of $EF$. Hence $XY$ is perpendicular bisector of $EF$ so we are done.

Solved with mananaban. We use the standard complex numbers setup. By the intersection formula, we have $$p = \frac{ac(b + d) - bd(a + c)}{ac - bd}$$and after some algebra we can also get $$\overline{p}=\frac{a+c-b-d}{ac-bd}.$$Then $$\frac{e-f}{m-n}=\frac{a + b - c - d + (cd - ab)\overline{p}}{a + d - b - c}=\frac{-(a + d - b - c)(ab + cd - ac - bd)}{(ac-bd)(a + d - b - c)} = \frac{ab + cd - ac - bd}{bd - ac}.$$It now suffices to check $$\frac{ab + cd - ac - bd}{bd - ac}+\frac{\frac{1}{ab} + \frac{1}{cd} - \frac{1}{ac} - \frac{1}{bd}}{\frac{1}{bd} - \frac{1}{ac}}=0,$$which is clear after a bit of simplification.

The idea here is that $MN$ is a very "bad" line in the sense that it's not easy to find angles involving it. Thus, we try to find a line parallel to $MN$. Note that because the are midpoints, we should think of reflections. Let $P'_1$ and $P'_2$ denote the reflections of $P$ across $M$ and $N$. Let $AB$ and $CD$ meet at $K$. By isogonality lemma on $\triangle KBC$ and points $A$ and $D$, $KP'_1$ is isogonal to $KP.$ Similarly, $KP'_2$ is also isogonal to $KP$, so $K,P'_1,KP'_2$ are collinear. Furthermore, $P'_1P'_2$ is parallel to $MN$ due to a homothety at $P$ with factor 2. Note that $PEKF$ is cyclic. Let line $KP'_1P'_2$ meet $EF$ at $R$. We then have $$\angle PKF=\angle PEF=\angle EKR,$$and since $\angle PEK=90$ we have $\angle KRE=90$ as well, done.

We claim that $ME=MF$ and analogously $NE=NF$, which suffices as this proves $MENF$ is a kite. Let $P$ and $Q$ be the midpoints of $\overline{BX}$ and $\overline{CX}$, respectively. Since $MP=QX=QF$ $PE=PX=MQ$ $\angle MPE=\angle MPX+2\angle EBX=\angle MQX+2\angle FCX=\angle MQF$, we have $PME \cong QFM$, so $ME=MF$, as desired. $\square$ Remark: Once we guess $ME=MF$, we can restate the problem as follows. Quote: In $\triangle XBC$, let $M$ be the midpoint of $\overline{BC}$. Let $E$ and $F$ be points on the circles with diameters $BX$ and $CX$, respectively, such that $\measuredangle EBX=\measuredangle XCF$ (in directed angles). Prove that $ME=MF$. This can be solved easily using moving points, but it also motivates the above synthetic solution by making it clear why the points $P$ and $Q$ are useful.

Set $(ABCD)$ be the unit circle as usual. Then, let $P=\overline{AC} \cap \overline{BD}$. Then, \[p=\frac{ac(b+d)-bd(a+c)}{ac-bd}\]Note that, \begin{align*} \overline{p} &= \frac{\frac{1}{ac}\left(\frac{1}{b}+\frac{1}{d} \right) - \frac{1}{bd}\left(\frac{1}{a}+\frac{1}{c} \right)}{\frac{1}{ac}-\frac{1}{bd}}\\ &= \frac{a-b-c+d}{ac-bd} \end{align*}Then, \begin{align*} e &= \frac{1}{2}\left( p+a+b-ab\overline{p} \right)\\ &= \frac{abc+acd-bcd-abd+a^2c-a^2b+ab^2-b^2d}{2(ac-bd)} \end{align*}Similarly, \[f = \frac{abc+acd-bcd-abd+ac^2-c^2d+cd^2-bd^2}{2(ac-bd)}\]Then, let $M_1,M_2$ be the midpoints of $AD$ and $BC$. This gives, \[m_1=\frac{a+d}{2} \text{ and } m_2=\frac{b+c}{2}\]Then, we observe that, \begin{align*} \frac{m_1-m_2}{e-f} &= \frac{\frac{a-b-c+d}{2}}{\frac{abc+acd-bcd-abd+a^2c-a^2b+ab^2-b^2d-(abc+acd-bcd-abd+ac^2-c^2d+cd^2-bd^2)}{2(ac-bd)}}\\ &= \frac{(a-b-c+d)(ac-bd)}{(a-b-c+d)(ac+bd-ab-cd)}\\ &= \frac{ac-bd}{ac+bd-ab-cd} \end{align*}We can then see that, \begin{align*} \overline{\left(\frac{m_1-m_2}{e-f} \right)} &= \frac{\frac{1}{ac}-\frac{1}{bd}}{\frac{1}{ac}+\frac{1}{bd}-\frac{1}{ab}-\frac{1}{cd}}\\ &= \frac{\frac{bd-ac}{abcd}}{\frac{bd+ac-cd-ab}{abcd}}\\ &= -\left(\frac{ac-bd}{ac+bd-ab-cd}\right) &= -\left(\frac{m_1-m_2}{e-f}\right) \end{align*}which implies that $\frac{m_1-m_2}{e-f}\in i\mathbb{R}$ and thus $EF \perp M_1M_2$ which was the required conclusion.

What are y'all waffling about? This is trivial by vectors! We use vectors, where we set $K=\overline{AC} \cap \overline{BD}$ to be the origin. It suffices to show that \[ ((\vec A - \vec B)+(\vec D - \vec C)) \cdot (\vec E - \vec F) = 0. \]By the definitions of $E$ and $F$ in the problem, we have \[ \vec E \cdot (\vec A - \vec B) = 0 \ \ \text{and} \ \ \vec F \cdot (\vec C - \vec D) = 0. \]Summing the above two equalities we obtain \[ \vec E \cdot (\vec A - \vec B) - \vec F \cdot (\vec D - \vec C) = 0. \]Thus it suffices to show that \[ \vec E \cdot (\vec D - \vec C) - \vec F \cdot (\vec A - \vec B) = 0. \]We do this synthetically. Let $\overline{KF}$ intersect $\overline{AB}$ at $F'$ and $\overline{KE}$ intersect $\overline{CD}$ at $E'$. Then \[ \vec E \cdot (\vec {CD}) - \vec F \cdot (\vec {BA}) = - KE \cdot CD \cdot \cos(\angle KE'D) + KF \cdot AB \cdot \cos(\angle KF'A). \]The fact that the RHS is zero follows by similar triangles on $\triangle AKB$ and $\triangle DKC$, and we conclude.

Let the intersection of the diagonals be $P$, and let the midpoints be $M$ and $N$. Reflect $P$ across $M$ and $N$ to points $Q$ and $R$, and let $X = \overline{AB} \cap \overline{CD}$. The isogonality lemma shows us that points $X$, $Q$, and $R$ lie on the isogonal to $\overline{XP}$ with respect to $\angle X$. It is well-known that the orthocenter and circumcenter of a triangle are isogonal conjugates, meaning that since $\overline{XP}$ is a diameter of $\triangle XEF$, its isogonal is the altitude, which means that $\overline{XST} \perp \overline{EF}$. Clearly, $\overline{MN} \parallel \overline{ST}$, so we are done. $\square$

Our solution is motivated by the first isogonality lemma. Reflect $X = AC \cap BD$ over $M$, $N$ to $P$, $Q$, respectively. If we let $Y = AB \cap CD$, we know $YPQ$ is isogonal to $YX$ in $\triangle YAD$. As $YX$ is the diameter of $(XEYF)$, its isogonal $YPQ$ is the altitude from $Y$ to $EF$. Homothety at $X$ then tells us \[EF \perp YP \parallel MN. \quad \blacksquare\]

Haha complex go brrrr. Denote by $M$ and $N$ the midpoints of $\overline{AD}$ and $\overline{BC}$. We set $(ABCD)$ as the unit circle. Compute, \begin{align*} p &= \frac{ac(b+d) - bd(a+c)}{ac - bd} \end{align*}Now we may also compute, \begin{align*} e &= \frac{1}{2}(a + b + p - ab\overline{p})\\ f &= \frac{1}{2}(c + d + p - cd\overline{p})\\ m &= \frac{1}{2}(a + d)\\ n &= \frac{1}{2}(b+c) \end{align*}Then we wish to show, \begin{align*} \frac{e - f}{m - n} \in i\mathbb{R} \end{align*}Scale all their coordinates by $2$ so that we wish to show, \begin{align*} \frac{a + b - c - d - \overline{p}(ab - cd)}{a - b - c + d} \in i\mathbb{R} \end{align*}The term, \begin{align*} \overline{p} &= \overline{\left(\frac{ac(b+d) - bd(a+c)}{ac-bd} \right)}\\ &= \frac{1/ac(1/b + 1/d) - 1/bd(1/a+1/c)}{1/ac - 1/bd}\\ &= \frac{-a + b - c +d}{bd - ac}\\ &= \frac{a - b + c - d}{ac - bd} \end{align*}Then we find, \begin{align*} \frac{a + b - c - d - \frac{(a - b + c - d)(ab - cd)}{ac - bd}}{a - b - c + d} \in i \mathbb{R}\\ \iff \frac{(ac - bd)(a+b-c-d) - (a - b + c - d)(ab - cd)}{(ac - bd)(a - b - c + d)} \in i \mathbb{R}\\ \iff \frac{(a - d) (c - b) (a - b - c + d)}{(ac - bd)(a - b - c + d)} \in i \mathbb{R}\\ \iff \frac{(a - d)(c-b)}{(ac - bd)} \in i \mathbb{R} \end{align*}which is easily checked to be true. $\square$

We employ complex numbers with $(ABCD)$ the unit circle. Note that for the intersection $$x = \frac{ac(b+d)-bd(a+c)}{ac-bd}$$we want $$\frac{f-e}{m-n} =\frac{c+d-a-b+\overline x(ab-cd)}{a+d-b-c}$$to be imaginary. In particular, note that $\overline x = \frac{a+c-b-d}{ac-bd}$, so \begin{align*} \frac{f-e}{m-n} &= \frac{(c+d-a-b)(ac-bd)+(a+c-b-d)(ab-cd)}{(a+d-b-c)(ac-bd)} \\ &= \frac{(a-d)(b-c)(a+d-b-c)}{(a+d-b-c)(ac-bd)} \\ &= \frac{(a-d)(b-c)}{ac-bd}. \end{align*}This expression clearly equals the negative of its conjugate, as needed.

Let $X$ be the intersection of diagonals $\overline{AC}$ and $\overline{BD}$, let $G$ be the intersection of lines $AB$ and $CD$ and let $K$ and $L$ and $M$ be the midpoints of $\overline{BC}$, $\overline{AD}$ and $\overline{XG}$ respectively. It's well known that $M$, $L$ and $K$ are collinear. Furthermore, since $GEXF$ is cyclic with center $M$, it follows that $ME = MF$. Claim: the line through the midpoints of $\overline{AD}$ and $\overline{BC}$ bisects $\overline{EF}$. Proof: Animate $E$ to slide from $A$ to $B$ in one second; animate $F$ to slide from $D$ to $C$ in one second. Because of coordinates, the midpoint of $\overline{EF}$ slides along a line. Since at $t=0$, this midpoint is $L$, and at $t=1$, this midpoint is $K$, it follows that the midpoint of $\overline{EF}$ always lies on $\overline{KLM}$. In particular, at some point in time, $E$ and $F$ are both the feet, which is what we care about. Since $M$ lies on the perpendicular bisector of $\overline{EF}$ and $\overline{KLM}$ bisects $\overline{EF}$, it follows that $\overline{KLM}$ is the perpendicular bisector of $\overline{EF}$.

I was literally this close to complex bashing... We first deal with when $AB \parallel CD,$ in which case the line connecting the midpoints is parallel to both $AB$ and $CD$ while $EF$ is clearly perpendicular to both. Henceforth assume that $AB \not \parallel CD.$ Now, we define some points. Let $M$ and $N$ be the midpoints of $AD$ and $BC$, respectively. Let $X = AC \cap BD$ and $Y = AB \cap CD.$ Reflect $X$ across $M$ and $N$ to get points $M'$ and $N'$ so that $MN \parallel M'N'.$ It suffices to show that $EF \perp M'N'.$ However, Claim: $Y, M', N'$ are collinear, and moreover their common is isogonal to $XY$ in $\angle BYC.$ Proof: Since $ABCD$ is cyclic, by the First Isogonality Lemma, we see that $YM'$ is isogonal to $YX$ in $\angle BYC.$ Repeat for $N',$ and we finish our proof of the claim. Thus it suffices to show that the isogonal line to $XY$ in $\angle EYF$ is perpendicular to $EF.$ However, since $EYFX$ is cyclic with $\angle YEX = \angle YFX = 90^\circ,$ we see that $YX$ passes through the circumcenter of $\triangle YEF,$ so the isogonal line of $YX$ in $\angle EYF$ is perpendicular to $EF,$ as desired.

If $AB \parallel CD$ then $ABCD$ is an isosceles trapezoid and the result is clear. Otherwise, let $X = AC \cap BD$ and $Y = AB \cap BC$. Let $M$ and $N$ denote the midpoints of $AD$ and $BC$ respectively. Reflect $X$ over $M$ and $N$ to $P$ and $Q$ respectively. From the First Isogonality Lemma, $YP$ and $YQ$ are both isogonal to $XY$ in $\angle BYC$, so $Y, P, Q$ are collinear. The circumcenter of $XEYF$ lies on $XY$, and since the orthocenter and the circumcenter are isogonal conjugates, $PQ \perp EF$. Then since $MN \parallel PQ$, we get that $MN \perp EF$.

(a similar exploit) Let Q be the intersection of BF and CE, prove that PQ is perpendicular to EF.

Solution: Let $E$ be the midpoint of $AA'$ and $BB'$, let $F$ be the midpoint of $CC'$ and $DD'$, and let $M$ and $N$ be the midpoints of $BC$ and $AD$. Then $APD'$ and $DPA'$ are congruent by SAS so $AD'=DA'$ so medians $NE$ and $NF$ must also have equal length. Thus $NEMF$ is a kite, and the result follows. Solution to Extension: Let $AB$ and $CD$ meet at $X$. Then $PEFX$ is cyclic with diameter $PX$. Also $E$ and $F$ and $B$ and $C$ are an isogonal pair in $\angle EPF$. Thus $PQ$ and $PX$ are isogonal in $\angle EPF$ and the result follows.

Let $P,Q,R,S$ be reflections of $X$ over $E,F$ and midpoints of $BC,AD$ respectively, and let $T=AB\cap CD$. By first isogonality lemma $T,R,S$ are collinear along $\ell$ and isogonal to $TX$. Now $\measuredangle(\ell,TP)=\measuredangle(\ell,TA)+\measuredangle XTA=\measuredangle DTA=\measuredangle DTX+\measuredangle(TD,\ell)=\measuredangle(TQ,\ell)$ so $\ell$ bisects $\angle PTQ$, since $TP=TX=TQ$ we get $RS$ perpendicularly bisects $PQ$, homothetying back gives the conclusion