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

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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Hope this will help you.....

Answer 2

Answer:

Step-by-step explanation:

In ∆ AOB, OD is the bisector of angle AOB

∴OA/OB =AD/DB---------------eq(1)

Theorem used here….. Angle bisector property

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

i.e.

∴AD*BE*CF = DB*EC*FA

Hope you mark the brainliest answer