Answer :

kvnmurty
In any triangle Sin A / Sin B = BC / AC

Ratio of Sine of angles = Ratio of sides opposite the angles.

Sin ABD / sin ADB  = AD / AB

Sin DBC / Sin BDC = DC / BC

Angle bisector : =>  angle ABD = angle DBC
and sin ADB = sin (180 - BDC) = sin BDC

Hence we have :  AD / AB = DC / BC

THis is the angle bisector theorem.

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Here's the answer 

Refer the attachment for the figure

Theorem : 
The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

Given:​ 
In ∆ ABC , side AD bisects Angle BAC such that B - D - C.

[tex]{\underline {\bf {To \: prove : }}}[/tex]

[tex]\sf{ \frac{AB}{AC} = \frac{BD}{DC}}[/tex] 

Construction :

Draw seg DE perpendicular to side AB and seg DN perpendicular to side AC.

Proof :

Point D lies on the bisector of Angle BAC
By angle bisector Theorem, DE = DF ... (1)

Now,
Ratio of areas of the two triangles is equal to the ratio of the products of their bases and corresponding heights.

[tex]\sf{\frac{A(ADB)}{A(ADC)} = \frac{ (AB \times DE)}{ (AC \times DF )}}[/tex]

From 1, 

[tex]\sf{\frac{A(ADB)}{A(ADC)} = \frac{ (AB )}{ (AC )...(2)}[/tex]

Also,∆ADB and ∆ADC have common vertex A 
and their bases BD and DC lie on the same line BC. So their heights are equal.

Area of triangles with equal heights are proportional to their corresponding bases.

[tex]{\sf{\frac{A( ∆ADB)}{A(∆ ADC)} = \frac{BD}{DC}} [/tex]

From 2 and 3 ,

We get 

[tex]{\sf{\frac{ AB }{AC} = \frac{BD}{DC}}[/tex]
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