Prove that the internal bisector of an angle of a triangle divides the opposite sides in the ratio of the sides containing the angle.


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Question Number 26858 by $@ty@m last updated on 30/Dec/17

$P r o v e t h a t t h e i n t e r n a l b i s e c t o r o f a n a n g l e$ $o f a t r i a n g l e d i v i d e s t h e o p p o s i t e$ $s i d e s i n t h e r a t i o o f t h e s i d e s c o n t a i n i n g$ $t h e a n g l e .$
	Answered by mrW1 last updated on 30/Dec/17

	Commented by mrW1 last updated on 30/Dec/17

	$\frac{B D}{\sin ∠ B A D} = \frac{A B}{\sin ∠ A D B}$ $\Rightarrow \frac{B D}{\sin α} = \frac{A B}{\sin β}$ $\Rightarrow \frac{B D}{A B} = \frac{\sin α}{\sin β}$ $\frac{D C}{\sin ∠ D A C} = \frac{A C}{\sin ∠ A D C}$ $\Rightarrow \frac{D C}{\sin α} = \frac{A C}{\sin (180 - β)} = \frac{A C}{\sin β}$ $\Rightarrow \frac{D C}{A C} = \frac{\sin α}{\sin β}$ $\Rightarrow \frac{B D}{A B} = \frac{D C}{A C} o r \frac{B D}{D C} = \frac{A B}{A C}$
	Commented by $@ty@m last updated on 31/Dec/17

	$T h a n k s . . .$
	Answered by mrW1 last updated on 31/Dec/17

	Commented by mrW1 last updated on 31/Dec/17

	$A n o t h e r e a s i e r w a y :$ $\frac{B D}{D C} = \frac{B E}{F C} = \frac{A B}{A C}$
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