In
a trapezium ABCD, AB || DC and DC = 2AB. EF || AB, where E and Flie on BC and AD respectively
such that BE/EC=4/3.Diagonal DB intersects EF at G. Prove that, 7EF = 11 AB.
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Answer:
given DC = 2 AB
Draw BH parallel to AD. H will be the middle point of DC as AB = DH.
Sin EF || AB, FI = AB
The ΔBHC and ΔBIE are similar as the corresponding sides are parallel.
So IE / HC = BE / BC
IE = HC * BE / BC
= y * 4x/7x = 4 y / 7
EF = FI + IE= y + 4 y /7 = 11 y / 7
so 7 EF = 11 AB
pleaseark as brilliant
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