4. [4] In a triangle $A B C$, take point $D$ on $B C$ such that $D B=14, D A=13, D C=4$, and the circumcircle of $A D B$ is congruent to the circumcircle of $A D C$. What is the area of triangle $A B C$ ?
Answer: 108
The fact that the two circumcircles are congruent means that the chord $A D$ must subtend the same angle in both circles. That is, $\angle A B C=\angle A C B$, so $A B C$ is isosceles. Drop the perpendicular $M$ from $A$ to $B C$; we know $M C=9$ and so $M D=5$ and by Pythagoras on $A M D, A M=12$. Therefore, the area of $A B C$ is $\frac{1}{2}(A M)(B C)=\frac{1}{2}(12)(18)=108$.