Question

property of an angle bisector of a triangle

Answer 1

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 

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

Answer 2

Answer:

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