Question 142
Consider a square $ABCD$ and the points $E$ on the side $CD$, $M$ on the diagonal $AC$ and $P$ on the side $BC$, such that $\angle BAE = \angle AEM = \angle AMP$. Prove that:
a) the triangle $AMP$ is isosceles;
b) $EM = DE + PB$.
a) the triangle $AMP$ is isosceles;
b) $EM = DE + PB$.