Question

O is any point in the interior of ∆ABC. 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

Answer 1

In ∆ AOB, OF is bisector of ∠ AOB

∴ OA / OB = AF / B ...... (1) (by angle bisector theorem)

In ∆ BOC, OD is bisector of angle ∠ BOC.

∴ OB / OC = BD / CD ....... (2)(by angle bisector theorem)

In ∆ AOC , OE is bisector of angle∠ AOC.

∴ OC / OA = CE / AE ....... (3)(by angle bisector theorem)

∴ BF × AE × CD = AF × CE × BD

Answer 2

By property of angle bisector

AF/BF=OA/OB

EC/AE=OC/OA

BD/DC=OB/OC

MULTIPLYING THE THREE EQNS WE GET

AF×EC×BD=BF×AE×CD

I am not sure if my answer is correct

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