Question

please prove
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

-----------------------------------------------------------------------------------------------------

AD*BE*CF = DB*EC*FA

---------------------------------------------------------------------------------------------------

Hope this will help you.....

Answer 2

Answer:

By the angle bisector theorem in triangle AOB, we have: AO/BO=AF/FB —-(1). Likewise, BO/CO=BD/CD ——(2) and CO/AO=CE/AE—-(3) . (1)x(2)x(3) gives us, (AO/BO)(BO/CO)(CO/AO)=(AF/FB)(BD/CD)(CE/AE) which implies,

BFxAExCD=AFxCExBD.

Step-by-step explanation: