Quadrilateral $ABCD$, whose diagonals are perpendicular to each other, is inscribed in a circle.

Perpendiculars dropped from vertices $B$ and $C$ to side $AD$ intersect diagonals $AC$ and $BD$ at points $E$ and $F$ respectively. Find $EF$, if $BC=1$.
Prove that BCFE is a parallelogram.

## Solution

First method.

Let $\angle C B D=\alpha$. Then

$$
\angle C A D=\alpha, \angle B E C=90^{\circ}-\alpha, \angle D B E=\alpha
$$

Therefore, $B E=B C$.

Let $\angle A C B=\beta$. Similarly, we can prove that $\angle A C F=\beta$. Therefore, $C F=B C$. Hence, $B E=C F$, and since $B E \| C F$ (perpendiculars to the same line $A D$), $B C F E$ is a parallelogram (even a rhombus). Therefore, $E F=$ $B C=1$.

Let $Q$ be the point of intersection of the diagonals. Note that $E$ is the point of intersection of the altitudes of triangle $A B D$. By a known property of the orthocenter of a triangle, $E Q=Q C$. Similarly, $F Q=Q B$. Since the diagonals of quadrilateral $B C F E$ are perpendicular and bisected by the point of intersection $Q$, $B C F E$ is a rhombus. Therefore, $E F=B C=1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-05.jpg?height=478&width=504&top_left_y=184&top_left_x=777)

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