Let $ABCD$ be a parallelogram with $\angle DAB < 90$ Let $E$ be the point on the line $BC$ such that $AE = AB$ and let $F$ be the point on the line $CD$ such that $AF = AD$. The circumcircle of the triangle $CEF$ intersects the line $AE$ again in $P$ and the line $AF$ again in $Q$. Let $X$ be the reflection of $P$ over the line $DE$ and $Y$ the reflection of $Q$ over the line $BF$. Prove that $A, X, Y$ lie on the same line.
Problem
Source: MEMO 2022 I3
Tags: geometry
haha-SuperDaivish
02.09.2022 16:52
Denote $I=\odot AFD\cap \odot ABE$,then we can easily get that $F,I,B$are collinear and$D,I,E$are collinear.
$\angle DP'E=\angle DPE=\angle DFA=\angle DCE$,so $D,P',C,E,A$are concyclic.
Then we get $\angle DAP'=\angle DEP'=\angle DEP=\angle IEA=\angle IBA=\angle DFI=\angle IAD$,which means that $A,I,P'$are collinear.
Likewise,$A,I,Q'$are collinear,so$A,P',Q'$are collinear.Done. $\Box$
Attachments:
keglesnit
02.09.2022 17:14
It is not hard to prove that $ADPF$ and $ABQE$ are isosceles trapezoids. Let the circumcircles of these quadrilaterals meet at $A$ and $Z$. Now $\angle AZE=\angle AQB=\angle AFD$ so $E$, $Z$, and $D$ must be colinear. Similarly, $F$, $Z$, and $B$ are colinear. Let the perpendicular line from $P$ onto $DE$ intersect $AZ$ at $X'$. Then $\angle DZX=\angle AFD=\angle PZD$ since $AF=AD=DP$. We conclude that $X=X'$. Similarly for $Y$ and that should finish the problem.
Tintarn
15.09.2022 02:12
Actually, we can completely do without circles! First of all remove the one existing circle by noting that equivlently $ADPF$ and $ABQE$ are isosceles trapezoids, in fact congruent ones. Also $ABF$ and $AED$ are congruent. Now $XD=PD=AF=AD$ and similarly $YB=AB$ and the rest is an angle chase: We have \[\angle YAB=90^\circ-\frac{1}{2} \angle ABY=90^\circ-\frac{1}{2} \angle ABF-\frac{1}{2} \angle FBY=90^\circ-\frac{1}{2}\angle ABF-\frac{1}{2} \angle QBF=\angle DAB-\angle ABF\]so that $\angle DAY=\angle ABF$ and similarly $\angle XAB=\angle EDA$ so that their sum is $180^\circ-\angle DAE=\angle DAB$ as desired.