Question 174
Let $ABC$ be a triangle, and let $M$ be the foot of the median from $A$. Let also $\ell_{\mathrm{b}}$ be the bisector of $\widehat{AMB}$ and $\ell_{c}$ the bisector of $\widehat{AMC}$. Finally, let $B'$ be the orthogonal projection of $B$ onto $\ell_{b}$, let $C'$ be the orthogonal projection of $C$ onto $\ell_{c}$, and let $A'$ be the intersection point of the lines $(AM)$ and $(B'C')$.
Prove that $A'B' = A'C'$.
Prove that $A'B' = A'C'$.