A right triangle $A B C$ is inscribed in a circle. A chord $C M$ is drawn from the vertex $C$ of the right angle, intersecting the hypotenuse at point $K$. Find the area of triangle $A B M$, if $A K: A B=1: 4, B C=\sqrt{2}, A C=$ 2.
Apply the theorem of the products of segments of intersecting chords.

## Solution

Let $\angle A B C=\beta$. From the right triangle $A B C$, we find that

$$
A B=\sqrt{A C^{2}+B C^{2}}=\sqrt{4+2}=\sqrt{6}, \cos \beta=\frac{B C}{A B}=\frac{1}{\sqrt{3}}
$$

Then $B K=\frac{3}{4} A B=\frac{3}{4} \sqrt{6}$. By the cosine theorem from triangle $B K C$, we find that

$$
C K=\sqrt{B C^{2}+B K^{2}-2 \cdot B C \cdot B K \cos \beta}=\sqrt{2+\frac{27}{8}-2 \cdot \sqrt{2} \cdot \frac{3}{4} \sqrt{6} \cdot \frac{1}{\sqrt{3}}}=\sqrt{\frac{19}{8}}
$$

By the theorem of the products of segments of intersecting chords, $C K \cdot K M=A K \cdot K B$, from which we find that

$$
K M=\frac{A K \cdot K B}{C K}=\frac{\frac{\sqrt{6}}{4} \cdot \frac{3 \sqrt{6}}{4}}{\sqrt{\frac{19}{8}}}=\frac{9 \sqrt{2}}{4 \sqrt{19}}
$$

Triangles $A B C$ and $A M B$ have a common base $A B$, so their areas are in the ratio of the heights dropped from vertices $C$ and $M$. The ratio of the specified heights is equal to the ratio of segments $C K$ and $K M$. Therefore,

$$
S_{\Delta \mathrm{ABM}}=\frac{K M}{C K} \cdot S_{\Delta \mathrm{ABC}}=\frac{\frac{9 \sqrt{2}}{4 \sqrt{19}}}{\sqrt{\frac{19}{8}}} \cdot \frac{1}{2} \cdot 2 \cdot \sqrt{2}=\frac{9}{19} \sqrt{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-28.jpg?height=620&width=688&top_left_y=-1&top_left_x=685)

## Answer

$\frac{9}{19} \sqrt{2}$.

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