Circle inscribed in Equilateral Triangles The circle inscribed in the triangle $ABC$ touches the sides $BC$ , $CA$ , and $AB$  in the points $A_1,B_1,C_1$ respectively. Similarly the circle inscribed in the triangle $A_1B_1C_1$ touches the sides in $A_2,B_2,C_2$ respectively, and so on. If $A_nB_nC_n$ be the $n^{th}$ $\triangle$ so formed, Prove its angles are: $$\frac{\pi}{3} + (-2)^{-n}(A - \frac{\pi}{3})\ ,\ \ \frac{\pi}{3} + (-2)^{-n}(B - \frac{\pi}{3})\ ,\ \ \frac{\pi}{3} + (-2)^{-n}(C - \frac{\pi}{3})$$
Hence, prove that triangle so formed is an equilateral triangle.

Quang Hoang has already provided a good hint.
Let us prove that by induction on $n$.
Let $O$ be the incenter of $\triangle{ABC}$. 
Then, noting that $OB\perp A_1C_1,OC\perp A_1B_1$, we have
$$\begin{align}\angle{B_1A_1C_1}&=\pi -\angle{B_1A_1C}-\angle{C_1A_1B}\\&=\pi-\left(\pi-\frac{\pi}{2}-\frac C2\right)-\left(\pi-\frac{\pi}{2}-\frac B2\right)\\&=\frac{B+C}{2}\\&=\frac{\pi-A}{2}\\&=\frac{\pi}{3}-2^{-1}\left(A-\frac{\pi}{3}\right)\end{align}$$
Similarly, we have
$$\angle{A_1B_1C_1}=\frac{\pi}{3}-2^{-1}\left(B-\frac{\pi}{3}\right),\qquad \angle{A_1C_1B_1}=\frac{\pi}{3}-2^{-1}\left(C-\frac{\pi}{3}\right)$$
Here, suppose that $$\angle{A_nB_nC_n}=\frac{\pi}{3}+(-2)^{-n}\left(B-\frac{\pi}{3}\right),\quad \angle{B_nC_nA_n}=\frac{\pi}{3}+(-2)^{-n}\left(C-\frac{\pi}{3}\right).$$
Let $O_n$ be the incenter of $\triangle{A_nB_nC_n}$.
Then, noting that $O_nB_n\perp A_{n+1}C_{n+1},O_nC_n\perp A_{n+1}B_{n+1}$, we have
$$\begin{align}&\angle{B_{n+1}A_{n+1}C_{n+1}}\\&=\pi -\angle{B_{n+1}A_{n+1}C_n}-\angle{C_{n+1}A_{n+1}B_n}\\&=\pi-\left(\frac{\pi}{2}-\frac 12\left(\frac{\pi}{3}+(-2)^{-n}\left(C-\frac{\pi}{3}\right)\right)\right)-\left(\frac{\pi}{2}-\frac 12\left(\frac{\pi}{3}+(-2)^{-n}\left(B-\frac{\pi}{3}\right)\right)\right)\\&=\frac{\pi}{3}+\frac{(-2)^{-n}}{2}(B+C)-(-2)^{-n}\cdot\frac{\pi}{3}\\&=\frac{\pi}{3}+\frac{(-2)^{-n}}{2}(\pi -A)-(-2)^{-n}\cdot\frac{\pi}{3}\\&=\frac{\pi}{3}+(-2)^{-n-1}\left(A-\frac{\pi}{3}\right)\end{align}$$
Similarly, we have
$$\small\angle{A_{n+1}B_{n+1}C_{n+1}}=\frac{\pi}{3}+(-2)^{-n-1}\left(B-\frac{\pi}{3}\right),\qquad \angle{A_{n+1}C_{n+1}B_{n+1}}=\frac{\pi}{3}+(-2)^{-n-1}\left(C-\frac{\pi}{3}\right)$$
