Problem:
Rectangle ABCD has AB=5 and BC=4. Point E lies on AB so that EB=1, point G lies on BC so that CG=1, and point F lies on CD so that DF=2. Segments AG and AC intersect EF at Q and P, respectively. What is the value of EFPQ​?
Answer Choices:
A. 163​​
B. 132​​
C. 829​
D. 9110​
E. 91​
Solution:
Triangles AEP and CFP are similar and FP:EP=CF : AE=3:4, so FP=73​EF. Extend AG and FC to meet at point H; then △AEQ and △HFQ are similar. Note that △HCG and △ABG are similar with sides in a ratio of 1:3, so CH=31​⋅5 and FH=3+35​=314​. Then FQ:EQ=314​:4=7:6, so FQ=137​FE. Thus PQ=FQ−FP=(137​−73​)FE=9110​FE and FEPQ​=(D)9110​​.
OR
Place the figure in the coordinate plane with D at the origin, A at (0,4), and C at (5,0). Then the equations of lines AC,AG, and EF are y=−54​x+4, y=−53​x+4, and y=2x−4, respectively. The intersections can be found by solving simultaneous linear equations: P(720​,712​) and Q(1340​,1328​). Because F,P, Q, and E are aligned, ratios of distances between these points are the same as ratios of the corresponding distances between their coordinates. Then
