The circle constructed on side $A C$ of triangle $A B C$ as a diameter passes through the midpoint of side $B C$ and intersects side $A B$ at point $D$ such that $A D=\frac{1}{3} A B$. Find the area of triangle $A B C$, if $A C=1$.

#
Triangle $A B C$ is isosceles.

## Solution

Let $M$ be the midpoint of $B C$. The diameter $A C$ is seen from point $M$ at a right angle, so $A M$ is the height and median of triangle $A B C$. Therefore, this triangle is isosceles, $A B=A C=1$. Thus, $A D=1 / 3, B D=2 / 3$, and since $B C \cdot B M=B A \cdot B D$, we have $2 B M^{2}=2 / 3, B M=\frac{1}{\sqrt{3}}$. Therefore,

$$
A M=\sqrt{A B^{2}-B M^{2}}=\sqrt{\frac{2}{3}} \cdot \text{ Therefore, } S_{A B C}=1 / 2 B C \cdot A M=\frac{\sqrt{2}}{3} .
$$