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 β³AED is twice the perimeter of β³BEC. 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 β³AED is isosceles. Segment EH is the common altitude of the two isosceles triangles β³AED and β³BEC, and
EH=102β62β=8.
Let AB=CD=x and AE=ED=y. Then 2x+2y+12=2(32), so y=26βx. Thus,