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?

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 $\alpha+\beta+\gamma=\pi$.
