Problem:
Quadrilateral ABCD is inscribed in circle O and has sides AB=3, BC=2,CD=6, and DA=8. Let X and Y be points on BD such that
BDDX=41 and BDBY=3611
Let E be the intersection of line AX and the line through Y parallel to AD. Let F be the intersection of line CX and the line through E parallel to AC. Let G be the point on circle O other than C that lies on line CX. What is XF⋅XG?
Answer Choices:
A. 17
B. 359−52
C. 491−123
D. 367−102
E. 18 Solution:
Because YE and EF are parallel to AD and AC, respectively, △XEY∼△XAD and △XEF∼△XAC. Therefore
XEXY=XAXD and XEXF=XAXC
It follows that
XDXC=XYXF
The Power of a Point Theorem applied to circle O and point X implies that XC⋅XG=XD⋅XB. Together with the previous equation this implies that XF⋅XG=XB⋅XY. Let d=BD; then DX=41d and BY=3611d. It follows that
