Given a square $A B C D$ with side length 1. Point $K$ lies on side $C D$ and $C K: K D=1: 2$. Find the distance from vertex $C$ to the line $A K$.

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Consider similar triangles.

## Solution

Let $P$ be the projection of vertex $C$ onto line $AK$. From the similarity of triangles $KRC$ and $KDA$, it follows that $CP: CK = AD: AK$. Therefore,

$CP = \frac{CK \cdot AD}{AK} = \frac{\frac{1}{3}}{\sqrt{1 + \left(\frac{2}{3}\right)^2}} = \frac{1}{\sqrt{13}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-35.jpg?height=363&width=548&top_left_y=785&top_left_x=755)

Answer

$\frac{1}{\sqrt{13}}$.

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