Question

O is any point in the interior of triangle ABC bisector of angleaob,angle boc and angle aoc intersect side ab, side bc ,side ac in f d e prove that bf x ae x cd =af x ce x bd ​

Answer 1

Step-by-step explanation:

hope it will help u.. the answer is in the pics..

Answer 2

Answer:

HEY MATE

BELOW IS YOUR ANSWER ✌✌

IN ABC

< AOB , < BOC AND < AOC INTERSECT BC , AC , AB IN F ,D, E

IN AOB

OF IS BISECTOR

AF / BF = OA / OB ..( BY ANGLE BISECTOR THEOREM )

AF x OB = BF x OA ..)(1#1

IN AOC

OE IS BISECTOR

BY ANGLE BISECOR THEOREM

AE / CE = OA/ OC

AE x OC = CE X OA

OA X CE = AE X OC .. (2)

IN BOC

OD IS BISECTOR

BY ANGLE BISECTOR THEOREM

BD / DC = OB/ DC

BD X OC = OB X DC ..(3)

FROM 1,2 ,3

=AF X OB X OA X CE XBD X OC = BF X OA X AE X OC X OB X DC

= AF X CE X BD = BF X AE X CD

= BF X AE X CD= AF X CE X BD

PLZZZZ

MARK ME AS A BRAINLIST