## [Thales' Theorem and the Theorem of Proportional Segments]

In parallelogram $A B C D$, diagonal $A C$ is twice the length of side $A B$. A point $K$ is chosen on side $B C$ such that $\angle K D B = \angle B D A$.

Find the ratio $B K: K C$.
Let $O$ - be the point of intersection of the diagonals of the parallelogram. From the condition, it follows that $A B=A O=O C=C D$ (see fig.).

![](https://cdn.mathpix.com/cropped/2024_05_06_dc8ed1d6cdac2bf77445g-38.jpg?height=354&width=604&top_left_y=1318&top_left_x=727)

Since $\angle K D B=\angle B D A=\angle D B K$, then $B K=K D$, so the median $K O$ of triangle $B K D$ is its altitude. Since $O C=C D$, the median $C Q$ of triangle $O C D$ is also its altitude. Thus, $C Q \| K O$, Therefore, $B K: K C=B O: O Q=2: 1$.

## Answer

$2: 1$.