<s>[INST] External bisectors of the angles of ABC triangle form a triangle $A_1B_1C_1$ and so on If the external bisectors of the angles of the triangle ABC form a triangle $A_1B_1C_1$,if the external bisectors of the angles of the triangle $A_1B_1C_1$ form a triangle $A_2B_2C_2$,and so on,show that the angle $A_n$ of the $n$th derived triangle is $\frac{\pi}{3}+(-\frac{1}{2})^n(A-\frac{\pi}{3})$,and that the triangles tend to become equilateral.
My attempt:The angle $A_1$ of triangle $A_1B_1C_1$ is $\frac{\pi}{2}-\frac{A}{2}$,angle $A_2$ of triangle $A_2B_2C_2$ is $\frac{A}{4}-\frac{3\pi}{4}$ but my answer is nowhere look like resembling final answer.Is my approach correct.If not,what is the correct way to solve this question?Can someone guide me?
 [/INST] You are on the right track, you just need to continue. You have a correct formula for $A_1$, so format it differently.
$$\begin{align}
A_1 &= \frac{\pi}2-\frac A2 \quad\text{(your formula)}\\
 &= \frac{\pi}3+\frac{\pi}6+\left(-\frac 12\right)A \\
 &= \frac{\pi}3+\left(-\frac 12\right)\left(A-\frac{\pi}3\right) \\
 &= \frac{\pi}3+\left(-\frac 12\right)^1\left(A-\frac{\pi}3\right) \\
\end{align}$$
Your formula for $A_1$ from $A$ also gives $A_{n+1}$ from $A_n$, so combining your formula with my derivation above and using induction we get
$$\begin{align}
A_{n+1} &= \frac{\pi}3+\left(-\frac 12\right)\left(A_n-\frac{\pi}3\right) \\
 &= \frac{\pi}3+\left(-\frac 12\right)\left[\frac{\pi}3+
   \left(-\frac 12\right)^n\left(A-\frac{\pi}3\right)-\frac{\pi}3\right] \\
 &= \frac{\pi}3+\left(-\frac 12\right)\left[
   \left(-\frac 12\right)^n\left(A-\frac{\pi}3\right)\right] \\
 &= \frac{\pi}3+\left(-\frac 12\right)^{n+1}\left(A-\frac{\pi}3\right) \\
\end{align}$$
which finishes the proof by induction.
So the given formula is correct. As we let $n\to\infty$ we see that $\left(-\frac 12\right)^n\to 0$ and thus $A_n\to\frac{\pi}3$. We can do the same to $B$ and $C$ to get $B_n\to\frac{\pi}3$ and $C_n\to\frac{\pi}3$, so all three angles approach $60°$, and the (increasingly large) triangles $\triangle A_nB_nC_n$ approach equilateral.

Here is a quick proof of your formula $A_1 = \frac{\pi}2-\frac A2$, though I used Greek letters for angles in this diagram.

This should be self-explanatory, and the final value for $A_1= \frac{\pi}2-\frac A2$ comes directly from $A_1=\frac{\beta+\gamma}2$ and $\alpha+\beta+\gamma=\pi$.
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