solve my question. ....
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Given Data: See figure:
O is point inside a triangle.
Bisectors of ∠ AOB, ∠BOC and ∠COA= OD, OE and OF
We need to prove that:
AD×BE×CF = BD×EC×FA
Now prove:
In ΔAOB bisector is ∠AOB is OD.
So, AD/BD = AO/BO.......(i) [Bisector Theorem]
In Δ BOC bisector of ∠BOC is OE.
So, BE/EC = BO/CO.......(ii)
In ΔCOA bisector of ∠COA is OF.
So, CF/FA = CO/AO........(iii)
Substituting equation (i),(ii) and (iii) we get:
⇒AD/BD×BE/CE × CF/FA = AO/BO × BO/CO × CO/AO
⇒AD/BD×BE/EC × CF/FA = 1
⇒AD×BE×CF = BD×EC×FA
Hence Proved
O is point inside a triangle.
Bisectors of ∠ AOB, ∠BOC and ∠COA= OD, OE and OF
We need to prove that:
AD×BE×CF = BD×EC×FA
Now prove:
In ΔAOB bisector is ∠AOB is OD.
So, AD/BD = AO/BO.......(i) [Bisector Theorem]
In Δ BOC bisector of ∠BOC is OE.
So, BE/EC = BO/CO.......(ii)
In ΔCOA bisector of ∠COA is OF.
So, CF/FA = CO/AO........(iii)
Substituting equation (i),(ii) and (iii) we get:
⇒AD/BD×BE/CE × CF/FA = AO/BO × BO/CO × CO/AO
⇒AD/BD×BE/EC × CF/FA = 1
⇒AD×BE×CF = BD×EC×FA
Hence Proved
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abhi178:
i know, but i wnat to your language , explain what ?????
angle bisector in a triangle that divides the oposite sides into two line segments
i think here some data missing , without any more detail we can't solve this
good , then now, use similarly concept how ??????
it divided two line segments with same propotion like the other two sides have........so we can use similarly concept
you write AD/BD = AO/BO
sorry , but i not satisfied
ok
abhi, durag solution is correct
thanks sir
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i hope this will usful to u
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