Here's a really weird way of doing it, in fact I don't know which technique did I use. , so high chance of getting it wrong (but I think it is correct). I will be really glad if somebody checks this. INAMO Shortlist 2015 G5 wrote: Let $ABC$ be an acute triangle. Suppose that circle $\Gamma_1$ has it's center on the side $AC$ and is tangent to the sides $AB$ and $BC$, and circle $\Gamma_2$ has it's center on the side $AB$ and is tangent to the sides $AC$ and $BC$. The circles $\Gamma_1$ and $ \Gamma_2$ intersect at two points $P$ and $Q$. Show that if $A, P, Q$ are collinear, then $AB = AC$. Solution:- Claim:- $\Gamma_1$ and $\Gamma_2$ are congruent. We will first deal with this problem to make the Claim clearer. ... wrote: If $\Gamma_1$ and $\Gamma_2$ are two circles with centers of $O_1,O_2$ respectively such that tangent from $O_1$ to $\Gamma_2$ and tangent from $O_2$ to $\Gamma_1$ intersect at $A$, now if $A$ lies on the Radical Axis of $\Gamma_1$ and $\Gamma_2$ then $\Gamma_1$ and $\Gamma_2$ are congruent. Let $\Gamma_1\cap\Gamma_2=\{P,Q\}$ Let the tangent from $O_1$ to $\Gamma_2$ be $l_1$ and the tangent from $O_2$ to $\Gamma_1$ be $l_2$. So we are given that $l_1\cap\l_2=A$ and $A\in PQ$. Now it is trivial to see that if $\Gamma_1$ and $\Gamma_2$ are congruent then $A\in PQ$. Now we will deal with the case when $\Gamma_1$ and $\Gamma_2$ are not congruent. We will move $O_2$ on line $O_1O_2$, let the new positions of $O_2$ be named as $O_2'$. Now we will construct a circle with radius $O_2'Q$. Clearly $P$ lies on this circle. Now note that the position of $A$ always changes during this change in the position of $O_2$. So now note that $Q$ remains fixed but the position of $A$ is not constant. So, the angle made by the lines $AQ$ and $O_1O_2$ always changes while changing the size of $\Gamma_2$. So this clearly implies $\Gamma_1$ and $\Gamma_2$ have to be congruent as if $\Gamma_1$ and $\Gamma_2$ would not have been congruent then $AQ$ would not have remained perpendicular to $O_1O_2$ anymore from our argument given above. Returning back to the main problem So it's easy to see that $AO_1=AO_2$. Now let the Tangency points made by $\Gamma_1$ and $\Gamma_2$ with $BC$ be $X,Y$ respectively. Now note that $\angle O_1XY=\angle O_2YC$, $O_1X=O_2Y$ and $\angle BO_2X=\angle BAQ=\angle QAC=\angle CO_2Y$. Hence by congruence we get that $O_2B=O_2C$. Hence, $$AB=AO_1+O_1B=AO_2+O_2C=AC.\blacksquare$$

It is clear that the centers of the circles will be the feet of respective angle bisectors on the opposite sides (and radii are the distances from centers to side $BC$). Since $A$ lies on the radical axis, its power wrt the two circles is the same. Using angle bisector theorem and similarity ($h$ is the length of the $A-altitude$) $$\left(\frac{bc}{a+c}\right) ^2 + \left(\frac{ah}{a+c}\right) ^2 = \left(\frac{bc}{a+b}\right) ^2 + \left(\frac{ah}{a+b}\right) ^2$$This yields $b=c$