9.3. 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 $B K: K C$.
Answer: $2: 1$.

Let $O-$ be the point of intersection of the diagonals of the parallelogram. From the problem statement, it follows that $A B=A O=O C=C D$ (see Fig. 9.3). Since $\angle K D B=\angle B D A=\angle D B K$, then $B K=K D$, therefore

![](https://cdn.mathpix.com/cropped/2024_05_06_8fea6e72207237693b93g-1.jpg?height=306&width=525&top_left_y=1958&top_left_x=1378)

Fig. 9.3 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, $C Q \| K O$, then, by the theorem of proportional segments (or from the similarity of triangles $B O K$ and $B Q C$), we get that $B K: K C=$ $=B O: O Q=2: 1$.

+ a complete and justified solution

it is only proven that triangles ВКD and ОСD are isosceles

- a particular case is considered or only the answer is given