Question

the internal bisector of angle of a triangle divides the opposite side internally in the ratio of sides containing the angle .prove it

Answer 1

The internal bisector of angle of a triangle divides the opposite side internally in the ratio of sides containing the angle. Prove it.

Proof.

To prove $\frac{BK}{KC} = \frac{AB}{AC}$ .

For $\Delta ABK$ : $\frac{BK}{\sin \angle BAK} = \frac{AK}{\sin \angle B}$ , (law of sines)

$$\frac {B K}{\sin \left(\frac {1}{2} \angle B A C\right)} = \frac {A K}{\sin \angle B}, \sin \left(\frac {1}{2} \angle B A C\right) = \frac {B K \sin \angle B}{A K}.$$

For $\Delta AKC$ : $\frac{KC}{\sin\angle KAC} = \frac{AK}{\sin\angle C}$ , (law of sines)

$$\frac {K C}{\sin \left(\frac {1}{2} \angle B A C\right)} = \frac {A K}{\sin \angle C}, \sin \left(\frac {1}{2} \angle B A C\right) = \frac {K C \sin \angle C}{A K}.$$

So $\frac{BK\sin\angle B}{AK} = \frac{KC\sin\angle C}{AK}$ and $\frac{BK}{KC} = \frac{\sin\angle C}{\sin\angle B}$ .

For $\Delta ABC$ : $\frac{AB}{\sin\angle C} = \frac{AC}{\sin\angle B}$ , (law of sines)

$$\frac {A B}{A C} = \frac {\sin \angle C}{\sin \angle B}.$$

Thus $\frac{BK}{KC} = \frac{\sin\angle C}{\sin\angle B}$ and $\frac{AB}{AC} = \frac{\sin\angle C}{\sin\angle B}$ .

Therefore $\frac{BK}{KC} = \frac{AB}{AC}$

Answer.

Hypothesis is proved.