Problem:
In triangle ABC, angle C is a right angle and CB>CA. Point D is located on BC so that angle CAD is twice angle DAB. If AC/AD=2/3, then CD/BD=m/n, where m and n are relatively prime positive integers. Find m+n.
Answer Choices:
A. 10
B. 14
C. 18
D. 22
E. 26
Solution:
Let E denote the point on BC for which AE bisects ∠CAD. Because the answer is not changed by a similarity transformation, we may assume that AC=25 and AD=35. Apply the Pythagorean Theorem to triangle ACD to obtain CD=5, then apply the Angle Bisector Theorem to triangle CAD to obtain CE=2 and ED=3. Let x=DB. Apply the Pythagorean Theorem to triangle ACE to obtain AE=24, then apply the Angle Bisector Theorem to triangle EAB to obtain AB=(x/3)24. Now apply the Pythagorean Theorem to triangle ABC to get
(25)2+(x+5)2=(3x24)2
from which it follows that x=9. Hence BD/DC=9/5, and m+n=14.
OR
Denote by a the measure of angle CAE. Let AC=2u, and AD=3u. It follows that CD=5u. We may assume BD=5. (Otherwise, we could simply modify the triangle with a similarity transformation.) Hence, the ratio CD/BD we seek is just u. Since cos2a=2/3, we have sina=1/6. Applying the Law of Sines in triangle ABD yields
