[ Diameter, main properties [Ratio of areas of similar triangles]

On the leg $B C$ of the right triangle $A B C$, a circle is constructed with $B C$ as its diameter, intersecting the hypotenuse $A B$ at point $K$. Find the area of triangle $B C K$, if $B C=a, C A=b$.

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Triangles СКВ and АСВ are similar. Find the similarity coefficient.

## Solution

Since point $K$ lies on the circle with diameter $BC$, then $CK \perp AB$. Triangles $CKB$ and $ACB$ are similar by two angles, the similarity coefficient is $\frac{CB}{AB}=\frac{a}{\sqrt{a^{2}+b^{2}}}$. Therefore,

$$
S_{\triangle \mathrm{BCK}}=\left(\frac{a}{\sqrt{a^{2}+b^{2}}}\right)^{2} \cdot S_{\triangle \mathrm{ACB}}=\frac{a^{2}}{a^{2}+b^{2}} \cdot \frac{1}{2} a b=\frac{a^{3} b}{2 a^{2}+2 b^{2}}
$$

Since point $K$ lies on the circle with diameter $BC$, then $CK \perp AB$. Triangles $CKB$ and $ACB$ are similar by two angles, the similarity coefficient is $\frac{CB}{AB}=\frac{a}{\sqrt{a^{2}+b^{2}}}$. Therefore,

$$
S_{\triangle \mathrm{BCK}}=\left(\frac{a}{\sqrt{a^{2}+b^{2}}}\right)^{2} \cdot S_{\Delta \mathrm{ACB}}=\frac{a^{2}}{a^{2}+b^{2}} \cdot \frac{1}{2} a b=\frac{a^{3} b}{2 a^{2}+2 b^{2}}
$$

Since point $K$ lies on the circle with diameter $BC$, then $CK \perp AB$. Triangles $CKB$ and $ACB$ are similar by two angles, the similarity coefficient is $\frac{CB}{AB}=\frac{a}{\sqrt{a^{2}+b^{2}}}$. Therefore,

$$
S_{\triangle \mathrm{BCK}}=\left(\frac{a}{\sqrt{a^{2}+b^{2}}}\right)^{2} \cdot S_{\Delta \mathrm{ACB}}=\frac{a^{2}}{a^{2}+b^{2}} \cdot \frac{1}{2} a b=\frac{a^{3} b}{2 a^{2}+2 b^{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-08.jpg?height=205&width=466&top_left_y=918&top_left_x=796)

Answer

$\frac{a^{3} b}{2 a^{2}+2 b^{2}}$