In parallellogram $ABCD$, on the arc $BC$ of the circumcircle $(ABC)$, not containing the point $A$, we take a point $P$ and on the $[AC$, we take a point $Q$ such that $\angle PBC= \angle CDQ$. Prove that $(APQ)$ is tangent to $AB$.
Problem
Source: Turkey Junior National Olympiad 2022 P4
Tags: geometry, circumcircle
BarisKoyuncu
23.12.2022 23:53
See that $\angle ABP=\angle QDA$ and $\angle APB=\angle ACB=\angle QAD$ implies $\triangle ABP\sim\triangle QDA$. Then, $\frac{|PA|}{|AQ|}=\frac{|AB|}{|DQ|}=\frac{|CD|}{|DQ|}$ and $\angle PAQ=\angle PBC=\angle CDQ$. Hence, $\triangle PAQ\sim\triangle CDQ$. Thus, $\angle AQP=\angle DQC=\angle DQA=\angle BAP$, as desired.
Invert around $A$. See that as $ABCD$ is a parallelogram, we have $\triangle AC^*D^*\overset{+}{\sim}\triangle B^*C^*A$. Also, the condition $\angle PBC= \angle CDQ$ implies $\angle AQ^*D^*=\angle B^*P^*A$. Hence, we find that $\triangle AC^*D^*\cup Q^*\sim\triangle B^*C^*A\cup P^*$. Yields, $\frac{|C^*Q^*|}{|Q^*A|}=\frac{|C^*P^*|}{|P^*B^*|}\Rightarrow P^*Q^*//AB^*$, as desired.
PROF65
24.12.2022 00:51
Let $(O),(O')$ the circles of $ABC,ADC$ since $\angle ABC= \angle CDA$ then they are congruent and are symmetric about $AC$ then the reflection $P'$ of $P$ in $AC$ is on $(O')$ and $A,P'$ and $Q$ are collinear More since have $\angle DCA=\angle BAC=\angle BAP + \angle PAC$ then $ \angle BAP=\angle DQC=\angle P'QC=\angle CQP$ ( the last because $P'$ and $P$ are symmetric about $AC$) thereforethe result follows .
bin_sherlo
24.12.2022 13:23
My solution from the exam. Let the intersection of $AP$ and $DC$ be $T.$ $ABPC$ is circumscribed so $\angle QDC= \angle PBC= \angle PAC$. Because of $\angle TAQ=\angle TDQ$ $ADQT$ is circumscribed. $\angle DTQ=\angle DAQ=\angle ACB$… (1) $\angle BCP= \angle BAT= \angle ATC$…(2) From (1) and (2) $\angle PTQ= \angle ACP$ so $CPTQ$ is cyclic. $\angle PTC= \angle PQC$. We proved that $\angle AQP=\angle BAP$.
X.Allaberdiyev
04.08.2023 11:03
By angle chasing we have $\angle PBC=\angle PAC=\angle CDQ$ and from this we can easily get $\angle BAP=\angle BCP=\angle AQD$ $(1)$. And we have $\angle BPA=\angle BCA=\angle QAD$ $(2)$. From $(1)$ and $(2)$ we conclude that $BAP$ and $DQA$ triangles are similar, which gives us $AQ/DQ=AP/AB$, then $AQ/DQ=AP/DC$, then $AQ/AP=DQ/DC$ $(3)$. And from angle chasing $\angle PBC=\angle PAQ=\angle CDQ$ $(4)$. From $(3)$ and $(4)$ we get that $PAQ$ and $CDQ$ triangles are similar, which gives us $\angle PQC=\angle PQA=\angle AQD=\angle BAP$, as desired.