Problem:
In triangle ABC, side AC and the perpendicular bisector of BC meet in point D, and BD bisects ∠ABC. If AD=9 and DC=7, what is the area of triangle ABD ?
Answer Choices:
A. 14
B. 21
C. 28
D. 145
E. 285 Solution:
By the angle-bisector theorem, BCAB=79. Let AB=9x and BC=7x, let m∠ABD=m∠CBD=θ, and let M be the midpoint of BC. Since M is on the perpendicular bisector of BC, we have BD=DC=7. Then
cosθ=727x=2x
Applying the Law of Cosines to △ABD yields
92=(9x)2+72−2(9x)(7)(2x)
from which x=4/3 and AB=12. Apply Heron's formula to obtain the area of triangle ABD as 14⋅2⋅5⋅7=145.
