We consider a square $ABCD$ and a point $E$ on the segment $CD$. The bisector of $\angle EAB$ cuts the segment $BC$ in $F$. Prove that $BF + DE = AE$.
Source: 2017 Pan-African Shortlist - G1
Tags: geometry, square, angle bisector
We consider a square $ABCD$ and a point $E$ on the segment $CD$. The bisector of $\angle EAB$ cuts the segment $BC$ in $F$. Prove that $BF + DE = AE$.