Question 101
Let $ABC$ be a given triangle. Let $D$ be a point on $[BC]$ such that $|AD| = \frac{|BD|^2}{|AB| + |AD|} = \frac{|CD|^2}{|AC| + |AD|}$ and $E$ be a point such that $D \in [AE]$ and $|CD| = \frac{|DE|^2}{|CD| + |CE|}$. Prove that $|AE| = |AB| + |AC|$. (Ali Doğanaksoy).