Let $ABCD$ be a cyclic quadrilateral. The diagonals $AC$ and $BD$ meet at $P$, and $DA $ and $CB$ meet at $Q$. Suppose $PQ$ is perpendicular to $AC$. Let $E$ be the midpoint of $AB$. Prove that $PE$ is perpendicular to $BC$.
Problem
Source: Pakistan TST(2) 2017.P1
Tags: cyclic quadrilateral, perpendicular lines, geometry
v_Enhance
29.01.2017 03:31
This was posted in the WOOT 1155 class forum two weeks ago. The hint I posted there was: Introduce $T = AA \cap BB$ and use Brokard's theorem to eliminate the points $D$ and $Q$ from the picture.
Forevermore
29.01.2017 04:17
Here's a solution that doesn't invoke any projective results. In the following proof, all angles are directed. Let the foot from $A$ onto $QB$ be $F$, so that $F$ lies on $\odot(APQ)$, and let $S = \overline{BP}\cap\overline{FA}$. We will show that $P$ is the center of $\odot(SFB)$. The result then follows by taking the homothety centered at $B$ with dilation factor $\tfrac{1}{2}$. Note that \begin{align*} 90^{\circ}-\measuredangle FSP &= \measuredangle SBF = \measuredangle SBC \\ &= \measuredangle DBC = \measuredangle DAC \\ &= \measuredangle QAC = \measuredangle QAP = \measuredangle QFP. \end{align*}Thus we have \[\measuredangle FSP = 90^{\circ}-\measuredangle QFP = 90^{\circ} - (90^{\circ}+\measuredangle AFP) = \measuredangle PFA = \measuredangle PFS, \]implying that $\triangle PFS$ is isosceles with $PF=FS$. But $P$ lies on $\overline{SB}$ and $\triangle SFB$ is right, thus $P$ is the circumcenter of $\triangle SFB$ as desired.
Aiscrim
01.02.2017 21:25
Here's a solution that doesn't invoke any brain. Let $T$ be the projection of $P$ onto $BC$ and let $\{M\}=PT\cap AB$. Denote $x=\widehat{ACB},\ y=\widehat{AQP}$. Note that $$\dfrac{TC}{TB}=\dfrac{PC}{PB} \cdot \dfrac{ \sin{\widehat{TPC}}}{ \sin{\widehat{TPB}}}=\dfrac{\cos{x}\cdot \cos{y}}{\sin{x}\cdot \sin{y}},\ \dfrac{PA}{PC}=\dfrac{QA}{QC}\cdot\dfrac{ \sin{\widehat{AQP}}}{ \sin{\widehat{CQP}}}=\dfrac{\sin{x}\cdot \sin{y}}{\cos{x}\cdot \cos{y}}$$Thus, by Menelaus in $\triangle{ABC}$, we get that $MA=MB$, whence the conclusion.
juckter
01.02.2017 21:49
Let $R$ be the reflection of $A$ about point $P$. Then $PQ \perp AC$ implies $QAR$ is isosceles. Now since $ABCD$ is cyclic we have $\angle QBP = \angle QAC = \angle QRA = \angle QRP$ and thus $QPRB$ is a cyclic quadrilateral. It follows that $RB$ is perpendicular to $BC$, hence $PE$ is too, as it is the midline opposite $A$ in triangle $ABR$.
bobthesmartypants
02.02.2017 03:13
Let $(QAC)\cap (QBD) = \{Q, Q'\}$. By radical axis theorem on $(ABCD), (QAC), (QBD)$, $Q,P,Q'$ are collinear. Since $A,B,C,D$ are concyclic, then $AC, BD$ are antiparallel w.r.t $Q$, so $\triangle QAC\sim \triangle QBD$. Since $QP$ is the $Q$-altitude of $\triangle QAC$, it must be the $Q$-diameter of $\triangle QBD$, so $Q'B\perp QC$. Let $H$ be the orthocenter of $\triangle QAC$. Then $AH\perp QC$ and since $Q'P=HP$, it follows that the distance between $Q'B$ and $P$ is the same as the distance between $AH$ and $P$. Thus, $PE \| AH\iff PE\perp BC$ as desired. $\Box$
TheCoolDinosuar
20.05.2022 21:37
Following vEnchance's hint, let $T=AA\cap BB$. Then $T$ lies on $PQ$ since the polar of $AB\cap CD$ is $PQ$ and $AB\cap CD$ lies on the polar of $T$ which is $AB$. Then we get the AIME II 2020/15 configuration. Let $P'$ be the projection of $T$ onto $CB$ and $M$ the midpoint of $AB$. Let the line through $T$ antiparallel to $AB$ wrt $\angle ACB$ intersect $CA$ and $CB$ at $E$ and $D$. Then $T$ is the center of cyclic quadrilateral $DABE$, so $P$ and $P'$ are the midpoint of $AD$ and $BE$. Then by Varignon parallelogram we get that $TPMP'$ is a parallelogram, and $PM\parallel TP'\perp $AB$ as desired.