Problem:
In â–³ABC shown in the figure, AB=7,BC=8,CA=9, and AH is an altitude. Points D and E lie on sides AC and AB, respectively, so that BD and CE are angle bisectors, intersecting AH at Q and P, respectively. What is PQ?
Answer Choices:
A. 1
B. 85​3​
C. 54​2​
D. 158​5​
E. 56​ Solution:
Let x=BH. Then CH=8−x and AH2=72−x2=92−(8−x)2, so x=2 and AH=45​. By the Angle Bisector Theorem in △ACH,PHAP​=CHCA​=69​, so AP=53​AH. Similarly, by the Angle Bisector Theorem in △ABH,QHAQ​=BHBA​=27​, so AQ=97​AH. Then PQ=AQ−AP=(97​−53​)AH=458​45​=158​5​​.
