1. Given is $\triangle A B C$, over whose side $A C$ a circle $k$ is constructed as if over a diameter. The circle $k$ passes through the midpoint of side $B C$, and intersects side $A B$ at point $D$ such that $A D=\frac{3}{2} D B$. If $A C=60$, calculate the area of $\triangle A B C$.
1. Let $N$ be the midpoint of side $BC$, and also the intersection of $k$ with $BC$. We have $\measuredangle ANC = \measuredangle ANB = 90^{\circ}$, because $AC$ is the diameter of circle $k$. By the SAS (Side-Angle-Side) theorem, we have $\triangle ANC \cong \triangle ANB$, so it follows that $AB = AC = 60$. Now we have $60 = AD + DB = \frac{3}{2} DB + DB = \frac{5}{2} DB$, so $DB = 24$ and $AD = \frac{3}{2} \cdot 24 = 36$. We also have $\measuredangle ADC = 90^{\circ}$ (again because $AC$ is the diameter), so now using the Pythagorean theorem we get $CD = \sqrt{60^2 - 36^2} = 12 \sqrt{5^2 - 3^2} = 48$. Since $CD$ is the height to $AB$ in $\triangle ABC$, its area is $\frac{AB \cdot CD}{2} = \frac{60 \cdot 48}{2} = 1440$.