Let $AD$ be the angle bisector of angle $A$ in $\Delta ABC$. Prove that $BD = BC$ . $\frac{AB}{AB + AC}$

Question

Let $AD$ be the angle bisector of angle $A$ in $\Delta ABC$. Prove that $$BD = BC \cdot \frac{AB}{AB + AC}$$

Hello,

I was doing some geometry and got stuck in this question. I tried using the angle bisector theorem and I know it will be used somewhere in this problem but can’t really get it right. Can you please help me with this question? I would be grateful if you did.

Thanks.

Answer 1

Using the theorem about the internal bisector, we get $$\frac{BD}{DC}=\frac{c}{b}$$ so we get $$BD=DC\cdot \frac{c}{b}$$ Using that $$DC=a-BD$$ We get $$BD=(a-BD)\cdot \frac{c}{b}$$ and we obtain $$BD(1+\frac{c}{b})=\frac{ac}{b}$$ Can you finish?

Answer 2

Extend side $AB$ across $A$ for $AC$, we get new point $E$. Then $\angle AEC = \angle BAD$ so $EC||AD$. By Thales theorem we obtain: $$ {BD\over BA}= {BC\over BE}\implies BD = {BA\cdot BC\over BE} ={AB\cdot BC\over AB+AC} $$