Problem: In right triangle ABC the hypotenuse AB=5 and leg AC=3. The bisector of angle A meets the opposite side in A1. A second right triangle PQR is then constructed with hypotenuse PQ=A1B and leg PR=A1C. If the bisector of angle P meets the opposite side in P1, the length of PP1 is:
Answer Choices:
A. 436
B. 435
C. 433
D. 232
E. 16152
Solution:
Let BA1=x∴4−xx=35,x=25 and 4−x=23.∴PR=4−x=23 and PQ=x=25.
∴△PQR∼△BAC, the ratio of the sides being 1:2. In △BAC,AA12=32+(23)2=445∴AA1=235 But PP1:AA1=1:2∴PP1=21⋅235=435.