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

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 \dots

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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