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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.

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  • $\begingroup$ What is the orgin of this problem? JEE? $\endgroup$ – N.S.JOHN Apr 20 '16 at 16:36
  • $\begingroup$ Trigonometry by Hall and Knight ,Loney are good old references in India. $\endgroup$ – Narasimham Apr 20 '16 at 16:47
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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: $$ 2(a+b+c) = (a+b)+(a+c)+(b+c).$$

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  • $\begingroup$ which editor or software you used to draw this picture, I am in the search of some user friendly tools through which I can draw complex geometrical shapes with ease. $\endgroup$ – user3290550 Nov 28 '19 at 14:32

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