Solution

a) Due to the metric relations in right triangles applied to triangle $A B P$, we have

$$
B E^{2}=A B^{2}=B F \cdot B P
$$

Therefore,

$$
\frac{B E}{B P}=\frac{B F}{B E}
$$

Given the proportional relationship of the last equation and $\angle E B F=\angle E B P$, it follows that triangles $B E F$ and $B E P$ are similar.

b) From the similarity in the previous item, it follows that $\angle B E F=\angle B P E$. Similarly, $\angle C E G=$ $\angle E P C$. Thus,

$$
\angle B E F+\angle C E G=\angle B P E+\angle E P C=\angle B P C=85^{\circ} \text {. }
$$
Let $A B C D$ be a rectangle with $B C=2 \cdot A B$. Let $E$ be the midpoint of $B C$ and $P$ an arbitrary point on the side $A D$. Let $F$ and $G$ be the feet of the perpendiculars drawn from $A$ to $B P$ and from $D$ to $C P$. We know that $\angle B P C=85^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_12f29789e10e6749ebb9g-22.jpg?height=402&width=717&top_left_y=1816&top_left_x=551)

a) Verify that the triangles $B E F$ and $B E P$ are similar.

b) Determine the value of the sum of the angles $\angle B E F+\angle C E G$.