The triangle angle bisector theorem was proved with the help of the sine rule.
Let us say
∠ABD = ∠CBD = α
∠ADB = β
So, ∠BDC = 180- β
What is the sine rule of a triangle?
In a triangle ABC, if the side in front of ∠A is a, in Infront of ∠B is b, in front of ∠C is c then, [tex]\frac{SinA}{a} =\frac{SinB}{b} =\frac{Sin}{c}[/tex].
In triangle ABD
From sine rule
Sin α /AD = Sin β /AB
Sin α /Sinβ = AD/AB...........(1)
In triangle BDC
From sine rule
Sin α / DC = Sin(180-β)/BC
as we know that,
Sin(180-β) = sin β
so, Sin α / DC = Sin β / BC
Sin α /Sinβ = DC/BC.........(2)
From equation (1) and (2)
[tex]\frac{AD}{AB} = \frac{DC}{BC} \\\\\frac{AB}{BC} =\frac{AD}{DC}[/tex]
Hence, the triangle angle bisector theorem was proved with the help of the sine rule.
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