# If $r$ is the inradius of $\triangle ABC$,then prove that $\frac{2}{r}=\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}$

In acute angled triangle $ABC$,a semicircle with radius $r_a$ is constructed with its base on $BC$ and tangent to the other two sides.$r_b$ and $r_c$ are defined similarly.If $r$ is the inradius of $\triangle ABC$,then prove that $\frac{2}{r}=\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}$

I am stuck at this problem.I drew it but i could not fully understand the problem.No worth mentioning inputs from me.Please help.

• What is the orgin of this problem? JEE? – N.S.JOHN Apr 20 '16 at 16:36
• Trigonometry by Hall and Knight ,Loney are good old references in India. – Narasimham Apr 20 '16 at 16:47

If a circle is tangent to $AB$ and $AC$, its centre lies on the bisector of $\widehat{BAC}$. If such a centre lies on $BC$, it is the feet $L_A$ of the angle bisector of $\widehat{BAC}$. By the bisector theorem: $$L_A B = \frac{ca}{b+c}$$ hence: $$r_A = L_A B \sin B = \frac{2\Delta}{b+c}.$$ On the other hand, it is trivial that $r = \frac{2\Delta}{a+b+c}$, hence the claim is equivalent to: $$2(a+b+c) = (a+b)+(a+c)+(b+c).$$