Bariz - Problem

Source: Francophone 2024, Junior P3

Tags: geometry, geometry proposed

Tintarn

04.04.2024 21:37

Let $ABC$ be an acute triangle with $AB<AC$ and let $O$ be its circumcenter. Let $D$ be a point on the segment $AC$ such that $AB=AD$. Let $E$ be the intersection of the line $AB$ with the perpendicular line to $AO$ through $D$. Let $F$ be the intersection of the perpendicular line to $OC$ through $C$ with the line parallel to $AC$ and passing through $E$. Finally, let the lines $CE$ and $DF$ intersect in $G$. Show that $AG$ and $BF$ are parallel.

Note $AO$ and $A$-altitude are isogonal in $\angle A$, so $DE$ and $BC$ are antiparallel, i.e. $BDCE$ cyclic, and since $AB = AD$, $BDCE$ is an isosceles trapezoid. Claim: $F \in (BDCE)$. Proof: $180^\circ - \angle EFC = \angle DCF = \angle ACB + \angle BCF = \angle C + \angle A = \angle EBC$. Claim: $ADGE$ cyclic. Proof: $\angle DGC = \angle GDE + \angle DEG = \angle FDE + \angle DEC = \angle FCE + \angle BCE = \angle FCB = \angle A = \angle DAE$. Hence $\angle GAD = \angle GED = \angle CDE = \angle BFE \implies \angle (GA, AD) = \angle (BF,FE)$, since $AD \parallel EF$, $AG \parallel BF$, as desired.