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In a cyclic quadrilateral $A B C D$, the diagonals of which intersect at point $K$, it is known that $A B=a, B K$ $=b, A K=c, C D=d$. Find $A C$.

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Triangle $A B K$ is similar to triangle $D C K$.

## Solution

From the similarity of triangles $A B K$ and $D C K$ (by two angles), it follows that $A B: C D = B K: C K$. Therefore, $C K = C D \cdot \frac{B K}{A B} = d \cdot \frac{b}{a}$.

Thus, $A C = A K + K C = c + d \cdot \frac{b}{a} = \frac{a c + b d}{a}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-08.jpg?height=480&width=483&top_left_y=1558&top_left_x=799)

## Answer

$\frac{a c + b d}{a}$.