Question

In trapezium ABCD , AB// DC and DC = 3AB. EF is drawn parallel to AB which cuts AD at F and BC at E such that AF/FD = 3/5. Diagonal DB intersects EF at G . Find the ratio of EF to AB.

Answer 1

Draw a line AH from A,  parallel to BC, to intersect CD at H, and to intersect EF at I.

So we see that ABCH is a parallelogram.  So HC = AB.
DH = DC = BC = 3 AB - AB =2 AB.

ΔAIF  is similar to ΔAHD, as  AI || AH, IF || HD and AF || AD.

AF / FD = 3/5 
=>   AF / AD = 3 / (3+5) = 3/8

So Ratios between two triangles:

AF / AD = FI / DH
=>  FI = 3/8 * DH = 3/4 * AB

EF = EI + IF = AB + 3/4 * AB

=>  EF / AB = 7/4