1979
Problem - 3310
Let $M$ be a point inside $\triangle{ABC}$. Draw $MA'\perp BC$, $MB'\perp CA$, and $MC'\perp AB$ such that $BA'=BC'$ and $CA'=CB'$. Prove $AB'=AC'$.
By # 3308 \begin{align}
MA'\perp BC&\implies A'B^2 - MB^2 = A'C^2 - MC^2\\
MB'\perp CA&\implies B'C^2 - MC^2 = B'A^2 - MA^2\\
MC'\perp AB&\implies C'A^2 - MA^2 = C'B^2 - MB^2
\end{align}
Adding these three equations together and noticing $BA'=BC'$ and $CA'=CB'$ lead to the desired conclusion.