A good geometry problem

By the Pythagorean Theorem, $BC = 35$ . Letting $BD = x$ we can use the angle bisector theorem on triangle $ABC$ to get $x/12 = (35 - x)/37$ , and solving gives $BD = 60/7$ and $DC=185/7$ .

The area of triangle $AGF$ is $10/3$ that of triangle $AEG$ , since they share a common side and angle, so the area of triangle $AGF$ is $10/13$ the area of triangle $AEF$ .

Since the area of a triangle is $\large\ \frac{ab\sin{C}}2$ , the area of $AEF$ is $525/37$ and the area of $AGF$ is $5250/481$ .

The area of triangle $ABD$ is $360/7$ , and the area of the entire triangle $ABC$ is $210$ . Subtracting the areas of $ABD$ and $AGF$ from $210$ and finding the closest integer gives $\boxed{148}$ as the answer.