Question

✌✌♥♥State the property of angle bisector of a triangle.

Draw figure and explain

Also , Prove it

Thanks! ♥♥✌✌

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.

${\underline {\bf {Given : }}}$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

$\huge \bf{ \purple{hey}}$
$\huge{ \mathfrak{ \red{here \: is \: answer}}}$

$\bf{for \: figure \: see \: \: in \: pic}$



${\underline {\bf {Given : }}}$Given:​ In ∆ ABC , side AD bisects Angle BAC such that B - D - C.

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

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

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, 

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

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 

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

$\huge \boxed{ \boxed{ \pink{hope \: it \: helps}}}$

$\huge{ \orange{thanks}}$