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:
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Toss on the Cartesian plane with A=(5,12),B=(0,0), and C=(14,0).
Then AD∥BC and AB∥CE by the trapezoid condition where D,E∈BP. Since PC=10, point P is 1510=32 of the way from C to A and is located at (8,8). Thus line BP has equation y=x. Since AD∥BC and BC is parallel to the ground, we know D has the same y-coordinate as A, except it'll also lie on the line y=x. Therefore, D=(12,12).
To find the location of point E, we need to find the intersection of y=x with a line parallel to AB passing through C. The slope of this line is the same as the slope of AB, or 512, and has equation y=512x−5168. The intersection of this line with y=x is (24,24). Therefore point E is located at (24,24).
The distance DE is equal to the distance between (12,12) and (24,24), which is (D) 122.
OR
Using Stewart's Theorem we find BP=82. From the similar triangles BPA∼DPC and BPC∼EPA, we have
