# In trapezium ABCD, AB||CD, DC=2AB. EF||AB cuts AD in F and BC in E. . DB intersects EF at G .PROVE THAT 7EF=10AB

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(1) GE/DC = BE/BC = 3/7 [ BE : EC = 3 : 4 (given), ==> BC =BE+EC = 7]

OR GE = (3/7)xDC = 2 x (3/7) x AB as DC = 2xAB (Given).

OR GE = (6/7)xAB

(2) DG/DB = CE/CB = 4/7

(3) In triangles DFG and DAB, FG is parallel to AB. Hence triangles DFG and DAB are similar. [For similar reasons as for triangles BGE and BDC]

Therefore FG/AB = DG/DB = 4/7 [from (2) above]

OR FG = (4/7)xAB

Hence EF = GE + FG = (6/7)xAB + (4/7)xAB = (10/7)xAB

==> 7EF = 10AB