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.
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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
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
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