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
please explain step by step
Answers
Answer:
Step-by-step explanation:
In ∆ AOB, OD is the bisector of angle AOB
OA/OB =AD/DB---------------eq(1)
Theorem used here
[The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle]
In ∆BOC .OE is the bisector of angle BOC
OB/OC = BE/EC---------eq(2)
In ∆COA, OF is the bisector of angle COA
OC/OA =CF/FA-----------eq(3)
Multiplying eq 1, 2, 3
(OA/OB) * (OB/OC) * (OC/OA) = (AD/DB) * (BE/EC) * (CF/FA)
1= (AD/DB) * (BE/EC) * (CF/FA)
DB*EC*FA = AD*BE*CF
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AD*BE*CF = DB*EC*FA
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