Question

how to prove angle bisector theorem

Answer 1

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.

Answer 2

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.

${\underline {\bf {To \: prove : }}}$

$\sf{ \frac{AB}{AC} = \frac{BD}{DC}}$ 

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.

$\sf{\frac{A(ADB)}{A(ADC)} = \frac{ (AB \times DE)}{ (AC \times DF )}}$

From 1, 

$\sf{\frac{A(ADB)}{A(ADC)} = \frac{ (AB )}{ (AC )...(2)}$

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.

${\sf{\frac{A( ∆ADB)}{A(∆ ADC)} = \frac{BD}{DC}}$

From 2 and 3 ,

We get 

${\sf{\frac{ AB }{AC} = \frac{BD}{DC}}$