I think there is the following relation, isn't it?? $sin\frac{A}{2}=\frac{r_a}{AI_a}$ where $r_a$ is the radious of the excircle!! Now if we substitute and use $r_a=p \cdot tg\frac{A}{2}$ I think we find that it is equilateral!

I believe we get some more cases-- not sure!!!

$\cos \frac A2=\frac {s}{AI_a}\Longleftrightarrow AI_a=\frac {s}{\cos \frac A2}$. Thus, $\frac {AI_a}{a}=\frac {BI_b}{b}=\frac {CI_c}{c}\Longleftrightarrow a\cos \frac A2=b\cos \frac B2=c\cos \frac C2\Longleftrightarrow $ $a^3(s-a)=b^3(s-b)=c^3(s-c)\Longleftrightarrow a=b=c.$

There are indeed "some more cases", what is because $a^3\left(s-a\right)=b^3\left(s-b\right)=c^3\left(s-c\right)$ does not imply a = b = c. Anyway, let me give a complete solution of the problem. Theorem 1. Let ABC be a triangle, and let $I_a$, $I_b$ and $I_c$ be the centers of its A-excircle, B-excircle and C-excircle, respectively. Then, we have $AI_a: BI_b: CI_c=a: b: c$ if and only if - either a = b = c, - or b = c and $a=\frac{1+\sqrt5}{2}b$, - or c = a and $b=\frac{1+\sqrt5}{2}c$, - or a = b and $c=\frac{1+\sqrt5}{2}a$. Proof of Theorem 1. First, let's consider an arbitrary triangle ABC. Let the A-excircle of triangle ABC touch the line CA at a point T. Then, since $I_a$ is the center of this A-excircle, $I_aT\perp CA$, so the triangle $I_aTA$ is right-angled at T; also, $\measuredangle I_aAT=\frac{A}{2}$ (since the point $I_a$, being the center of the A-excircle of triangle ABC, lies on the angle bisector of the angle CAB), and AT = s, where $s= \frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. Thus, in the right-angled triangle $I_aTA$, we have $AI_a=\frac{AT}{\cos\measuredangle I_aAT}$, so that $AI_a=\frac{s}{\cos\frac{A}{2}}$. But by the half-angle formulas, $\cos\frac{A}{2}=\sqrt{\frac{s\left(s-a\right)}{bc}}$. Thus, $AI_a=\frac{s}{\sqrt{\frac{s\left(s-a\right)}{bc}}}=\sqrt{\frac{sabc}{a\left(s-a\right)}}$. Similarly, $BI_b=\sqrt{\frac{sabc}{b\left(s-b\right)}}$ and $CI_c=\sqrt{\frac{sabc}{c\left(s-c\right)}}$. Hence, we can equivalently transform the relation $AI_a: BI_b: CI_c=a: b: c$ as follows: $AI_a: BI_b: CI_c=a: b: c$ $\Longleftrightarrow\ \ \ \ \ \sqrt{\frac{sabc}{a\left(s-a\right)}}: \sqrt{\frac{sabc}{b\left(s-b\right)}}: \sqrt{\frac{sabc}{c\left(s-c\right)}}=a: b: c$ $\Longleftrightarrow\ \ \ \ \ \sqrt{\frac{1}{a\left(s-a\right)}}: \sqrt{\frac{1}{b\left(s-b\right)}}: \sqrt{\frac{1}{c\left(s-c\right)}}=a: b: c$ $\Longleftrightarrow\ \ \ \ \ \frac{1}{a\left(s-a\right)}: \frac{1}{b\left(s-b\right)}: \frac{1}{c\left(s-c\right)}=a^2: b^2: c^2$ $\Longleftrightarrow\ \ \ \ \ 1: 1: 1=\left(a^3\left(s-a\right)\right): \left(b^3\left(s-b\right)\right): \left(c^3\left(s-c\right)\right)$ $\Longleftrightarrow\ \ \ \ \ a^3\left(s-a\right)=b^3\left(s-b\right)=c^3\left(s-c\right)$. Since $s-a=\frac{a+b+c}{2}-a=\frac{b+c-a}{2}$ and similarly $s-b=\frac{c+a-b}{2}$ and $s-c=\frac{a+b-c}{2}$, this becomes $a^3\left(b+c-a\right)=b^3\left(c+a-b\right)=c^3\left(a+b-c\right)$. Now, let ABC be a triangle satisfying the condition $AI_a: BI_b: CI_c=a: b: c$. Then, we must therefore have $a^3\left(b+c-a\right)=b^3\left(c+a-b\right)=c^3\left(a+b-c\right)$. Now, since our situation is symmetric, we can WLOG assume that $a\geq b\geq c$. Since $b^3\left(c+a-b\right)=c^3\left(a+b-c\right)$, we have $0=b^3\left(c+a-b\right)-c^3\left(a+b-c\right)=a\left(b^3-c^3\right)-\left(b^4-c^4-b^3c+c^3b\right)$ $=a\left(b^2+bc+c^2\right)\left(b-c\right)-\left(b^3+c^3\right)\left(b-c\right)$ $=\left(a\left(b^2+bc+c^2\right)-\left(b^3+c^3\right)\right)\left(b-c\right)$. Now, $a\left(b^2+bc+c^2\right)-\left(b^3+c^3\right)=\underbrace{\left(a-b\right)}_{\geq 0\text{, since }a\geq b}b^2+\underbrace{\left(a-c\right)}_{\geq 0\text{, since }a\geq c}c^2+\underbrace{abc}_{>0}>0$; thus, we must have b - c = 0, so that b = c. Hence, the equation $a^3\left(b+c-a\right)=b^3\left(c+a-b\right)$ becomes $a^3\left(b+b-a\right)=b^3\left(b+a-b\right)$, i. e. $a^3\left(2b-a\right)=b^3a$. Division by a transforms this into $a^2\left(2b-a\right)=b^3$, and thus $0=a^2\left(2b-a\right)-b^3=\left(a-b\right)\left(ab-a^2+b^2\right)$. Hence, either a - b = 0, what leads to a = b and thus to a = b = c, or $ab-a^2+b^2=0$, what is a quadratic equation in a and has the solutions $a=\frac{1+\sqrt5}{2}b$ and $a=\frac{1-\sqrt5}{2}b$, of which the second one can be excluded since it is negative (and sidelengths of a triangle cannot be negative), so we get the solution b = c and $a=\frac{1+\sqrt5}{2}b$. The other solutions can be obtained similarly (we WLOG assumed that $a\geq b\geq c$ and thus didn't get the cyclic permutations). By tracing the above argumentation backwards, we see that these solutions are indeed solutions. Theorem 1 is proven. Darij

Darij, you are true. I was negligently (it is regrettably !) I will prove that at least two between $a,b,c$ are equally, i.e. $(a-b)(b-c)(c-a)=0$. Thus, $a\ne b\ne c\ne a\ \wedge \ a^3(b+c-a)=b^3(c+a-b)=c^3(a+b-c)\Longrightarrow$ $a^3+b^3=c(a^2+ab+b^2),\ b^3+c^3=a(b^2+bc+c^2),\ c^3+a^3=b(c^2+ca+a^2)\Longrightarrow$ $a^2+b^2+c^2=0$, what is absurd. Therefore, $a=b\ \vee\ b=c\ \vee\ c=a$ a.s.o.

Ya , darij! I got the same answers... Those were the other cases I was talking about above!