Asked by abhishek11 | 1st Feb, 2010, 10:58: AM
construction: from E, draw EY parallel to AC meeting BC in Y.
In triangle ADC,
EY II AC
DE/EA=DY/YC...(i).. by BPT
but DE/EA=1... given
in triangle XBC
EY II XC, since EY is parallel to AC by construction.
BE/EX=BY/YC=(BD+DY)/YC=(DC+YC)/YC=(2YC+YC)/YC=3... since D is mid pt of BC.
Answered by | 1st Feb, 2010, 06:08: PM
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