Let $ABCD$ be a trapezoid with bases $AB$ and $CD$. Bisectors of $AD$ and $BC$ intersect line segments $BC$ and $AD$ respectively in points $P$ and $Q$. Show that $\angle APD = \angle BQC$.
Problem
Source: XIII Polish Junior MO 2018 First Round - Problem 4
Tags: geometry, trapezoid, bisector, Poland
Vrangr
24.04.2018 17:10
$ABPQ$ and $CDPQ$ are cyclic and the result follows, I'm on mobile so I'll post complete proof later. Edit [27-04-2018]: Took me surprisingly long to use a desktop computer. Claim 1: $P, M, N, Q$ are concyclic. \[\measuredangle PMQ = 90^{\circ} = \measuredangle PNQ\] Claim 2: $A, B, P, Q$ and $C, D, P, Q$ are concyclic. \[\measuredangle QPB = \measuredangle QPM = \measuredangle QNM = \measuredangle QAB\]Similarly, \[\measuredangle QPC = \measuredangle QDC\] Main Proof: Let $X \in AD: PX \parallel AB \parallel CD$. \[\angle BQC = \angle BQP + \angle PQC = \angle BAP + \angle PDC = \angle DPX + \angle XPA = \angle DPA\]
WolfusA
26.04.2018 17:02
Let $M,N$ be the midpoints of segments $BC,AD$. $\angle PMQ=\angle PNQ=90^\circ\implies P,M,N,Q$ are concyclic. Hence $\angle CMN=\angle DQP$ regardless of point configuration. $MN ||AB\implies \angle PBA=\angle CMN =\angle DQP =180^{\circ} -\angle AQP$. Points $A,B$ lie on the same side of line $PQ$ hence quadrilateral $ABPQ$ is cyclic. So $AP = DP \wedge BQ = CQ\wedge \angle PAD=\angle QBC\implies $triangles $APD,BQC$ are similar. Conclusion: $\angle APD = \angle BQC$
Jzhang21
07.08.2018 18:24
Let $E, F$ be midpoints of $AD, BC$ respectively and let $\angle DAP=\angle ADP =x, \angle BCQ=\angle CBQ=y,$ and $\angle BAP=z.$ Because triangles $APD$ and $BQC$ are isosceles, it suffices to prove $\angle PDA=\angle PCQ or x=y$. Since $AB\parallel EF\parallel CD, \angle FED=\angle CDQ=x+z$. Because $\angle QEP=\angle QFP=90, QEFP$ is cyclic so $\angle FEP=\angle FED-\angle PED=x+z-90=\angle FQP$. Note that $\angle FQC=90-y$ so $\angle PQC=180-x-y-z.$ Thus, $\angle QPC=180-\angle PQC-\angle PCQ=x+z=\angle QDC$ so $QDPC$ is cyclic, implying $\angle PCQ=y=x=\angle PDA$ as desired. $\blacksquare$
AlastorMoody
21.11.2018 10:45
Let $M,N$ be mid-points of $AD,BC$ respectively, Since, $\angle QMP= \angle QNP =90^{\circ} \implies MNPQ \text{ is cyclic, and Since,} MN||AB \implies \angle QPB= 180^{\circ}- \angle QMN= 180^{\circ}-\angle QAB $ and Similarly, $\angle AQP =180^{\circ} - \angle ABP \implies ABPQ \text{ is cyclic }$, Hence, $\angle AQB=\angle ABP$, but since, $\Delta APD$ and $\Delta BQC$ are isosceles $\implies \boxed{\angle APD=\angle BQC}$