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
BDDXβ=41β and BDBYβ=3611β
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 XFβ XG?
Answer Choices:
A. 17
B. 359β52ββ
C. 491β123ββ
D. 367β102ββ
E. 18 Solution:
Because YE and EF are parallel to AD and AC, respectively, β³XEYβΌβ³XAD and β³XEFβΌβ³XAC. Therefore
XEXYβ=XAXDβ and XEXFβ=XAXCβ
It follows that
XDXCβ=XYXFβ
The Power of a Point Theorem applied to circle O and point X implies that XCβ XG=XDβ XB. Together with the previous equation this implies that XFβ XG=XBβ XY. Let d=BD; then DX=41βd and BY=3611βd. It follows that
To determine d, note that because ABCD is a cyclic quadrilateral it follows that Ξ±=β BAD=Οββ DCB. Applying the Law of Cosines to β³ABD and β³BCD yields
