A circle with center $O$ has an inscribed trapezoid $K L M N$, where $K N \| L M, K N=6, L M=4$, and the angle $K L M$ is $135^{\circ}$. Point $A$ lies on segment $K N$, such that $A K=4$. Find the distance from point $O$ to line $B N$.
Find $A L$ and consider similar triangles $K A L$ and $B A N$.

## Answer

$\frac{8}{\sqrt{5}}$

Points $C$ and $D$ lie on a circle with diameter $A B$ and are distinct from $A$ and $B$. Lines $A C$ and $B D$ intersect at point $P$, and lines $A D$ and $B C$ intersect at point $Q$. Prove that $A B$ is perpendicular to $P Q$.

## Hint

The altitudes of a triangle intersect at one point.

## Solution

Since points $C$ and $D$ lie on a circle with diameter $A B$ and are distinct from $A$ and $B$, then

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

so two altitudes of triangle $A B P$ lie on the lines $A D$ and $B C$, which intersect at point $Q$. Therefore, the line containing the third altitude $P M$ of this triangle also passes through point $Q$. Consequently, $P Q \perp A B$.
![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-15.jpg?height=488&width=1338&top_left_y=802&top_left_x=356)

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