Question

what is the angle bisector Theorem proof it now with the help of well labelled figure

Answer 1

Heya User,

--> Ummm, I'm sorry but I'm =_= soooo uncomfortable with Constructions.. So, I'll use a rather happy Geometry proof ^_^ 

--> Also, I use the notation [ ABC ] to define the area of ΔABC

--> Now, in ΔABC :->
    --> [ ADB ] / [ ADC ] = BD / DC  --> (i)

--> Note :-> [ Both triangles have a common height ]

Further, DE and DF are heights to AB and AC ..
--> Note :-> AD = AD || ∠EAD = ∠FAD || ∠AED = ∠AFD 

=> ΔAED is congruent to ΔAFD --> { ASA Criteria }
=> DE = DF --> { c.p.c.t }

Further, [ ABD ] / [ ACD ] = { 1/2 * AB * DE } / { 1/2 * AC * DF }
==> [ ABD ] / [ ACD ] = AB / AC --> (ii) 

Equating (i) and (ii) -->
---> AB / AC = BD / DC  Proved ....

Well, since u've asked to prove the angle bisector theorem,
     ^_^  I presume that you know what it states :p

--> Plz let me know if uh wish to ask anything ^_^

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 

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