2002 AMC 12A Problem 23

Since $D$ lies on the perpendicular bisector of $\overline{BC},$ $DB = DC = 7.$ The angle bisector $\overline{BD}$ gives $\dfrac{AB}{BC} = \dfrac{AD}{DC} = \dfrac{9}{7},$ so write $AB = 9x$ and $BC = 7x.$ Let $\theta = \angle ABD = \angle DBC.$ In isosceles $\triangle BDC,$ the foot of the perpendicular is the midpoint $M$ of $\overline{BC},$ so $\cos\theta = \dfrac{BM}{BD} = \dfrac{\frac{7x}{2}}{7} = \dfrac{x}{2}.$ Applying the Law of Cosines in $\triangle ABD:$ $$9^2 = (9x)^2 + 7^2 - 2(9x)(7)\cdot\dfrac{x}{2},$$ which simplifies to $81 = 18x^2 + 49,$ so $x = \dfrac43$ and $AB = 12.$ Now $\triangle ABD$ has sides $9, 7, 12.$ By Heron’s formula with $s = 14,$ the area is $\sqrt{14\cdot 5\cdot 7\cdot 2} = \sqrt{980} = 14\sqrt5.$ Thus, the correct answer is D.