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Diameter, main properties

On the leg $B C$ of the right triangle $A B C$ as a diameter, a circle is constructed, intersecting the hypotenuse at point $D$ such that $A D: B D=1: 3$. The height dropped from the vertex $C$ of the right angle to the hypotenuse is 3. Find the leg $B C$.

#
$C D$ - the height of triangle $A C D$.

## Solution

Since angle $B D C$ is inscribed in the given circle and rests on its diameter $B C$, then $\angle B D C=90^{\circ}$. Therefore, $C D$ is the height of triangle $A B C$.

Let $A D=x, B D=3 x$. Since $C D^{2}=A D \cdot D B$, then $3 x^{2}=9$. From this, we find that $x^{2}=3$. Therefore,

$$
B C^{2}=C D^{2}+B D^{2}=9+9 x^{2}=9+27=36
$$

and $B C=6$.

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-30.jpg?height=420&width=483&top_left_y=373&top_left_x=793)

Answer

6.