Question 77
Let $ABC$ be a triangle, isosceles at $A$, all of whose angles are acute, and let $D$ be a point on the segment $[BC]$. Let $\ell$ be the line parallel to $(BC)$ passing through $A$, then $X$ the point of $\ell$ for which $(XD)$ is perpendicular to $(BC)$, and $\Gamma$ the circle with centre $X$ passing through $D$. The circle $\Gamma$ meets the segment $[AB]$ at a point $E$, and we denote by $Y$ the point of $\ell$ such that $(AB)$ is perpendicular to $(EY)$; similarly, $\Gamma$ meets the segment $[AC]$ at a point $F$, and we denote by $Z$ the point of $\ell$ such that $(AC)$ is perpendicular to $(FZ)$.
Prove that the bisectors of the angles $\widehat{EYA}$ and $\widehat{FZA}$ meet at a point of the line $(XD)$.
Prove that the bisectors of the angles $\widehat{EYA}$ and $\widehat{FZA}$ meet at a point of the line $(XD)$.