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
Answers
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:
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: