In a triangle ABC a straight line is drawn from point E of side BC parallel to the altitude BD and cuts the side AC at a point F. A line segment EF divide the triangle into two figures of same area Then find the length of EF if BD =6cm and AD/DC=2/7
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Draw the diagram.Let AD=2x,DC=7x,EF=y.BD=6.BCD is similar to EFC.FE/BD=FC/CD OR y/6=FC/7x OR FC=7xy/6.[EFC]=[ABEF]=0.5[ABC] So,0.5XFCXEF=0.25XACXBD OR 0.5X7xy/6Xy=0.25X9xX6 OR 0.5X7(y^2)/6=0.25X9X6 OR y^2=23.14285714 OR y=4.810702354.So EF=y=4.810702354.