Bariz - Problem

Source: P3 Francophone Math Olympiad Senior 2023

Tags: geometry, tangent circles

parmenides51

02.05.2023 14:44

Let $ABCD$ be a convex quadrilateral, with $\measuredangle ABC > 90^\circ$, $\measuredangle CDA > 90^\circ$ and $\measuredangle DAB = \measuredangle BCD$. Let $E$, $F$ and $G$ be the reflections of $A$ with respect to the lines $BC$, $CD$ and $DB$. Finally, let the line $BD$ meet the line segment $AE$ at a point $K$, and the line segment $AF$ at a point $L$. Prove that the circumcircles of the triangles $BEK$ and $DFL$ are tangent to each other at $G$.

Note that $\angle AFD=\angle LAF = \angle DGL\Rightarrow D,G,F,L$are cyclic $\angle BGK=\angle BAE = \angle AEB \Rightarrow E,B,G,K$are cyclic Consider that $\angle KAL = \angle KAB+\angle A+\angle DAL=180^{\circ} - \angle BAC - \angle ACB - \angle CAD - \angle ACD + \angle A = 180-\angle A$ Let $l$ be the line tangent to $\odot (BKG)$ at $G$ Since $\measuredangle (l,KG) + \angle KLG = \angle A = \measuredangle (l,KG) + \measuredangle (l,GL)\Rightarrow \measuredangle(l,GL)=\angle KLG$ Hence$l$is the line tangent to $\odot (DGF)$ at $G$$\Rightarrow$$\odot(BEK)$and$\odot(DFL)$ are tangent to each other at $G$