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Question
O is any point in the interior of ΔABC. Bisectors of ∠AOB, ∠BOC and ∠AOC intersect side AB, side BC, side AC in
F, D and E respectively.
Prove that
BF × AE × CD = AF × CE × BD
Solution
In Δ AOB, OF is bisector of ∠ AOB
∴`(OA)/(OB) = (AF)/(BF)` ....... (1) (by angle bisector theoerm)
In Δ BOC, OD is bisector of angle ∠ BOC .
∴`(OB)/(OC) =( BD)/(CD)` ....... (2)(by angle bisector theoerm)
In Δ AOC , OE is bisector of angle∠ AOC.
∴`( OC)/(OA) = (CE)/(AE)` ....... (3)(by angle bisector theoerm)
∴ `(OA)/(OB) ×( OB)/(OC) × (OC)/(OA) = (AF)/(BF) ×( BD)/(CD) × (CE)/(AE)` from (1), (2) and (3)
`therefore(OAxxOCxxOB)/(OBxxOAxxOC) = (AFxxBFxxCE)/(BDxxAExxCD)`
` therefore1 = (AFxxCExxBD)/(BFxxAExxCD)`
∴ BF × AE × CD = AF × CE × BD
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