Let $ABCD$ be a trapezoid with bases $AB$ and $CD$. Points $P$ and $Q$ lie on diagonals $AC$ and $BD$, respectively and $\angle APD = \angle BQC$. Prove that $\angle AQD = \angle BPC$.
Problem
Source: XIII Polish Junior MO 2018 Second Round - Problem 4
Tags: geometry, trapezoid, Plane Geometry, Poland
georgeado17
24.04.2018 16:59
We have that $DPQC$ is cyclic which follows that $\angle DCA=\angle DQP=\angle CAB$ $\implies$ $APQB$ is cyclic with $\angle APB=\angle AQB$ and respectively $\angle BPC=\angle AQD$ done.
quangminhltv99
24.04.2018 18:12
[asy][asy]
size(250);
defaultpen(fontsize(8pt));
pair A = (-2,0);
pair B = (2,0);
pair D = (-1.3,1.5);
pair C = (1,1.5);
pair M = midpoint(A--D);
pair N = midpoint(B--C);
pair Pp = bisectorpoint(D,A);
pair Qp = bisectorpoint(C,B);
pair P = extension(M,Pp,B,C);
pair Q = extension(N,Qp,A,D);
draw(A--B--C--D--cycle^^B--P^^A--Q, blue);
draw(M--P^^N--Q, orange);
draw(A--P--D^^B--Q--C, magenta);
draw(M--N^^P--Q, purple);
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(N));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
[/asy][/asy]
Clearly, $\angle QMP=\angle QNP=90^{\circ}$, hence $MNPQ$ is cyclic. Thus, $\angle PQM+\angle NMQ=180^{\circ}$. On the other hand, we have $MN\parallel CD$ since $MN$ is midline of $ABCD$. Thus, $\angle CDQ+\angle PQM=\angle MNQ+\angle PQM=180^{\circ}$, or $CDPQ$ is cyclic. By the properties of cyclic quadrilateral, we have $\angle CPD=\angle CQD$.
Moreover, we have $MN\parallel AB$, any by similar arguments, we have $ABPQ$ is cyclic, and from there we conclude $\angle APB=\angle BQA$.
Thus, $\angle APD=\angle APB-\angle DPC=\angle BQA-\angle CQD=\angle BQC$.
Ari004
15.06.2018 10:30
The angle condition implies $DQPC$ cyclic,hence $\sphericalangle BCP=$ $\sphericalangle BQP$.Since $AB \parallel CD$,we have $\sphericalangle DCP=$ $\sphericalangle BAP$,implying $ABQP$ concyclic,hence done,
Jzhang21
12.08.2018 05:53
Since $\angle BQC=\angle APD$, then $\angle DQC=\angle DPC$ so $QPCD$ is cyclic. Hence, $\angle ABQ=\angle QDC=\angle QPA$ so $ABPQ$ is cyclic. $\angle QAD+\angle ADQ$ $=\angle A -(\angle QAP+\angle CAB)+\angle D-(\angle BDC)$ $=180^{\circ}-(\angle QAP+\angle PAB+\angle QDC)$ $=180^{\circ}-(\angle QBP+\angle PCD+\angle QBA)$ $=\angle B-(\angle QBP+\angle QBA)+\angle C -\angle PCD$ $=\angle PBC+\angle PCB.$ Subtracting both sides from $180^{\circ}$ gives $\angle AQD=\angle BPC,$ as desired. $\blacksquare$
MarkBcc168
12.08.2018 05:58
The angle condition implies $DQPC$ is cyclic so by Reim theorem, $ABPQ$ is cyclic too hence we are done.
Albert123
29.10.2021 20:06
Note that $DQPC$ is cyclic($\angle DQC=\angle DPC$) $\implies ABPQ$ is cyclic Let $\angle DCP=\angle PQA=\angle APB=\alpha$ $\implies \angle AQB=\angle APB$ Then:$180- \angle AQB=180-\angle APB$ $\implies \angle AQD=\angle BPC$
Elnuramrv
30.10.2021 08:27
Let $ \angle BQC = \angle APD = \alpha $ Then $ \angle DQC = \angle DPC = 180-\alpha \Rightarrow \Box DQPC cyclic \Rightarrow \angle QDC = \angle QPA = \beta $ $ AB // DC \Rightarrow \angle BDC = \angle ABD = \beta $ Now $ \angle ABQ = \angle AQP = \beta \Rightarrow ABPQ cyclic $ Then $ \angle AQB = \angle APB = \gamma $ Whence $ \angle AQD = \angle BPC = 180 - \alpha - \beta - \gamma$