In a trapezium ABCD, AB || DC and DC = 2AB. If EF is drawn parallel to AB cuts AD in F and BC in E such that BE / BC = ¾. Diagonals DB intersects EF at G. Find x for 7EF = 10AB
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As per figure, in triangles BGE and BDC, GE and DC are parallel, hence angle BGE = angle BDC and angle BEG = angle BCD, hence triangles BGE and BDC are similar.
(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
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