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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. |
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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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