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
Answers
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:
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) x (OB/OC) x (OC/OA) = (AD/DB) x (BE/EC) x (CF/FA)
1= (AD/DB) x (BE/EC) x (CF/FA)
DBxECxFA = ADxBExCF
-----------------------------------------------------------------------------------------------------
ADxBExCF = DBxECxFA