Question

O is a point inside a triangle ABC. The bisector of CAON, HOC ond ZCOA meet desides AB.
BC and CA in points and respectively. Prove that AD.BE.CF is DB.EC.FA​

Answer 1

Step-by-step explanation:

ANSWER

In △AOD, OD is the bisector of ∠AOB.

∴                OBOA=DBAD........(i)

In △BOC, OE is the bisector of ∠BOC.

∴                OCOB=ECBE......(ii)

In △COA, OF is the bisector of ∠COA.

∴               OAOC=FACF.......(iii)

Multiplying the corresponding sides of (i), (ii) and (iii), we get

          OBOA×OCOB×OAOC=DBAD×ECBE×FACF

⇒     1=DBAD×ECBE×FACF

⇒     DB×EC×FA=AD×BE×CF

⇒     A

Answer 2

Answer:

In ∆ AOB, since OD is the bisector of ∠ AOB.

∴ AD / DB = OA / OB -----> Equation 1

In ∆ BOC, since OE is the bisector of ∠ BOC.

∴ BE / EC = OB / OC -----> Equation 2

In ∆ COA, since OF is the bisector of ∠ COA.

∴ CF / FA = OC / OA ------> Equation 3

Multiply 1,2 and 3,we get

=> AD / DB × BE / EC × CF / FA = 1

=> AD / DB × BE / EC × CF / FA = 1

=> AD × BE × CF = DB × EC × FA.

Hence, the result proved.

Step-by-step explanation:

@Genius