Problem:
In the diagram below, angle ABC is a right angle. Point D is on BC, and AD bisects angle CAB. Points E and F are on AB and AC, respectively, so that AE=3 and AF=10. Given that EB=9 and FC=27, find the integer closest to the area of quadrilateral DCFG.
Solution:
By the Angle-Bisector Theorem, BD:DC=AB:AC=12:37, and thus the area of triangle ADC is 37/49 of the area of triangle ABC. By the Angle-Bisector Theorem, EG:GF=AE:AF=3:10, and thus the area of triangle AGF is 10/13 of the area of triangle AEF. The area of triangle AEF is 3/12 of the area of triangle AFB, which is in turn 10/37 of the area of triangle ABC. Since BC=372−122​=35, the area of triangle ABC is 210. It follows that the area of quadrilateral DCFG is
