[Inscribed angle, based on the diameter] [Pythagorean theorem (direct and inverse).]

In triangle $ABC$, $AB = BC = 6$. A circle is constructed on side $AB$ as the diameter, intersecting side $BC$ at point $D$ such that $BD: DC = 2: 1$.

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

By the Pythagorean theorem, $A D^{2}=A B^{2}-B D^{2}=20, A C^{2}=A D^{2}+D C^{2}=24$.

![](https://cdn.mathpix.com/cropped/2024_05_06_845c09db4774ba28be39g-36.jpg?height=400&width=397&top_left_y=2058&top_left_x=836)

## Answer

$2 \sqrt{6}$