Problem:
Triangle ABC has AB=13,BC=14 and AC=15. Let P be the point on AC such that PC=10. There are exactly two points D and E on line BP such that quadrilaterals ABCD and ABCE are trapezoids. What is the distance DE ?
Answer Choices:
A. 542
B. 62
C. 584
D. 122
E. 18
Solution:
Answer (D): Without loss of generality, suppose that D is closer to P than is E, and let PD=x. Note that AP=AC−PC=15−10=5 and △APD∼△CPB; therefore BP=x⋅510=2x. Because △APB∼△CPE, it follows that EP=2BP=4x. Hence DE=EP−DP=3x. The Law of Cosines in △ACB gives
132=142+152−2⋅14⋅15⋅cos∠ACB,
whence cos∠ACB=53. Then the Law of Cosines in △PCB gives
BP2=102+142−2⋅10⋅14⋅53=102+142−14⋅12=128.
Therefore BP=82,x=42, and DE=3⋅42=122.
OR
As in the first solution, BP=2x and EP=4x, so EB=6x. Furthermore, EC=AB⋅510=26. Using Stewart's Theorem on △CEB with base EB=6x and cevian CP=10 gives