[ Theorem P  [ Auxiliary

In trapezoid $A B C D$, one base is twice the length of the other.

The smaller base is equal to $c$.

The diagonals of the trapezoid intersect at a right angle, and the ratio of the non-parallel sides is $k$. Find the non-parallel sides of the trapezoid.

#
Express the lateral sides of the trapezoid in terms of the squares of the segments into which the diagonals of the trapezoid are divided by their point of intersection.

## Solution

Let $O$ be the point of intersection of the diagonals $A C$ and $B D$ of trapezoid $A B C D, B C=c, A D=2 c$.

Let $C D=x$. Then $A B=$ $k x$. Let $O C=a, O B=b$. Then from the similarity of triangles $A O D$ and $C O B$ it follows that $A O=2 a, O D=$ $2 b$.

By the Pythagorean theorem $A B^{2}=O B^{2}+$ $O A^{2}, C D^{2}=O C^{2}+O D^{2}$, or $k^{2} x^{2}=$ $b^{2}+4 a^{2}, x^{2}=a^{2}+4 b^{2}$.

Adding these two equations, we get $x^{2}\left(k^{2}+1\right)=5\left(a^{2}+b^{2}\right)$. From the right triangle $B O C$ it follows that $a^{2}+b^{2}=c^{2}$. Therefore, $x^{2}\left(k^{2}+1\right)=5 c^{2}$. Hence,

$C D=x=\frac{\sqrt{5} c}{\sqrt{1+k^{2}}}, A B=k x=\frac{\sqrt{5} k c}{\sqrt{1+k^{2}}}$

## Answer

$$
\frac{\sqrt{5} c}{\sqrt{1+k^{2}}}, \frac{\sqrt{5} k c}{\sqrt{1+k^{2}}}.
$$

## Problem