Question 102
Let $ABCD$ be a rectangle. Let $E$ be the foot of perpendicular from $A$ to $BD$. Let $F$ be an arbitrary point of the diagonal $BD$ between $D$ and $E$. Let $G$ be the intersection of the line $CF$ with the perpendicular from $B$ to $AF$. Let $H$ be the intersection of the line $BC$ with the perpendicular from $G$ to $BD$. Prove that $\measuredangle EGB = \measuredangle EHB$.