Problem:
Points A,B,C, and D lie on a line, in that order, with AB=CD and BC=12. Point E is not on the line, and BE=CE=10. The perimeter of â–³AED is twice the perimeter of â–³BEC. Find AB.
Answer Choices:
A. 15/2
B. 8
C. 17/2
D. 9
E. 19/2
Solution:
Let H be the midpoint of BC. Then EH is the perpendicular bisector of AD, and â–³AED is isosceles. Segment EH is the common altitude of the two isosceles triangles â–³AED and â–³BEC, and
EH=102−62​=8.
Let AB=CD=x and AE=ED=y. Then 2x+2y+12=2(32), so y=26−x. Thus,
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