[Isosceles, Inscribed, and Circumscribed Trapezoids] [Pythagorean Theorem (Direct and Inverse)]

In an isosceles trapezoid \(ABCD\) with the larger base \(AD=12\) and \(AB=6\). Find the distance from point \(O\) where the diagonals intersect to point \(K\) where the extensions of the lateral sides intersect, if the extensions of the lateral sides intersect at a right angle.
If $P$ is the midpoint of the base $A D$, then $K O=K P-O P$.

## Solution

Let $P$ be the midpoint of the base $A D$. Then $K P$ is the height of triangle $A K D$. Points $O$ and $M$ (the midpoint of the base $B C$) lie on the segment $K P$. Since

$\angle K A P=45^{\circ}$, then $K P=A P=6, K B=A K-A B=6(\sqrt{2}-1)$,

$B C=K B \sqrt{2}=6 \sqrt{2}(\sqrt{2}-1), \quad \frac{O P}{O M}=\frac{A D}{B C}=\frac{\sqrt{2}}{\sqrt{2}-1}$.

Therefore, $O P=\frac{M P \sqrt{2}}{\sqrt{2}-1+\sqrt{2}}=\frac{6}{2 \sqrt{2}-1}=\frac{6(2 \sqrt{2}+1)}{7}$,

$K O=K P-O P=6-\frac{6(2 \sqrt{2}+1)}{7}=\frac{12(3-\sqrt{2})}{7 \mathbf{K}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-30.jpg?height=392&width=594&top_left_y=1839&top_left_x=732)

## Answer

$\frac{12(3-\sqrt{2})}{7}$