Problem:
Let △ABC be a scalene triangle. Point P lies on BC so that AP bisects ∠BAC. The line through B perpendicular to AP intersects the line through A parallel to BC at point D. Suppose BP=2 and PC=3. What is AD?
Answer Choices:
A. 8
B. 9
C. 10
D. 11
E. 12
Solution:
Let Q be the intersection point of AC and BD, as shown.
Because BQ​⊥AP and ∠BAP=∠QAP, it follows that △ABQ is isosceles with AB=AQ. Then by the Angle Bisector Theorem,
23​=PBPC​=ABAC​=AQAC​=1+AQCQ​=1+ADBC​=1+AD5​.
Solving this equation yields AD=10.
OR
Let M be the intersection point of AP and BQ​. As above, △ABQ is isosceles with AB=AQ, so BM=MQ. Let BM=MQ=x,QD=y, and AD=z. Because △AMD and △PMB are similar, it follows that
2z​=xx+y​=1+xy​.
Further, △AQD and △CQB are also similar, so 5z​=2xy​. It follows that
2z​−1=xy​=52z​
which yields z=(C)10​.
The problems on this page are the property of the MAA's American Mathematics Competitions
