If $I_1$ and $I_2$ and $I_3$ be the centres of the escribed circles of $\triangle ABC$ and if $R_1$, $R_2$ and $R_3$ are radius of the circles inscribed in the triangles $\triangle BI_1C , \triangle AI_2C, \triangle AI_3B $, then prove that
$$R_1 : R_2 : R_3 =\sin(A/2):\sin(B/2):\sin(C/2).$$
I got $$ R_3 = 8R\sin(C/2) \cos(A/2)\cos(B/2)\cos(C/2) $$ But I am not sure whether it is correct or not.
Can anyone help me?
On
Let $R$ be the circumradius of $\Delta ABC$ and $r_a, r_b, r_c$ be the ex-radii of $\Delta ABC$ opposite to vertices $A, B, C$. Next, it is easy to observe that $$\Delta I_{A}BC \sim \Delta AI_{B}C \sim \Delta ABI_{C}.$$ $$\therefore \boxed{\frac{R_1}{I_{A}B}=\frac{R_2}{AI_{B}}=\frac{R_3}{AB}}. \;\;\; {(1)}$$ Now, note that \begin{align*} I_{A}B &= \frac{r_a}{\cos{\frac{B}{2}}} \\ &= \frac{4R\sin{\frac{A}{2}}\cos{\frac{B}{2}}\cos{\frac{C}{2}}}{\cos{\frac{B}{2}}} \\ &= 4R\sin{\frac{A}{2}}\cos{\frac{C}{2}}. \end{align*} Similarly, we have \begin{align*} AI_{B}&= \frac{r_b}{\cos{\frac{A}{2}}} \\ &= \frac{4R\cos{\frac{A}{2}}\sin{\frac{B}{2}}\cos{\frac{C}{2}}}{\cos{\frac{A}{2}}} \\ &= 4R\sin{\frac{B}{2}}\cos{\frac{C}{2}}. \end{align*} And the extended Law of Sines yields us $$ AB = c = 2R\sin{C} = 4R\sin{\frac{C}{2}}\cos{\frac{C}{2}}.$$ Putting these into $(1)$, we obtain $$\frac{R_1}{4R\sin{\frac{A}{2}}\cos{\frac{C}{2}}}=\frac{R_2}{4R\sin{\frac{B}{2}}\cos{\frac{C}{2}}}=\frac{R_3}{4R\sin{\frac{C}{2}}\cos{\frac{C}{2}}}.$$ $$\text{i.e.} \;\; \boxed{\frac{R_1}{\sin{\frac{A}{2}}}=\frac{R_2}{\sin{\frac{B}{2}}}=\frac{R_3}{\sin{\frac{C}{2}}}}.$$ $QED.$
I think there is a mistake in the question. The given result does not hold for the inscribed circles in those triangles but for the circumcircles.
It's really simple actually. Just draw $\triangle ABC$ and all its exterior and interior angle bisectors. Let the exterior ones meet at $I_1,I_2$ and $I_3$ . Doing some angle chasing you'll find that $\angle BI_1C=90^{\circ}-\frac A2$. Now if $R_1$ is the circumradius of $\triangle BI_1C$ then according to the sine law:$$R_1=\frac{BC}{2\sin\angle BI_1C}=\frac{a}{2\cos \frac A2}$$.
Also we know that $a=2R\sin A$ so just apply that in that to finally get $R_1=2R\sin \frac A 2$. After that there's not much left to do.