1.

The side BC of a `DeltaABC` is bisected at D. O is any point on AD. BO and CO are produced to meet AC and AB at E & F respectively. AD is produced to C1 so that D is the mid-point of OC1. Prove that FE||BC.

Answer» `square BC_1CO`
BC and `OC_1` are bisecting each other
`square BC_1CO` is a parallelogram
`BC_1||OC and C C_1||BO`
`BC_1||FC and C C_1||BE`
`C C_1||OE`
In`/_ABC_1`
`OF||BC_1`(proved)
`(AF)/(FB)=(AD)/(OC_1)`(Proporsnality)
In`/_AC C_1`(Proved)
`(AE)/(EC)=(AF)/(FB)`
FE||BC
(By P. Theorem).


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