Points $X,Y$ are located on the line segment $AB,AC$ respectively so that $BX.AB=IB^2$ and $CY.AC=IC^2$ A triangle $ABC$ has incenter $I$.Points $X,Y$ are located on the line segment $AB,AC$ respectively so that $BX.AB=IB^2$ and $CY.AC=IC^2$.Given that $X,I,Y$ are collinear,find the value of measure of angle $A.$

I could not solve this problem.I have no idea how to start tackling it.Please help me.Thanks.
 A: The given ($BX.AB = IB^2$) merely is used as the converse of power of a point. That is to say there is a red circle (centered at $O_R$) passing through A, X, and I with BI being the tangent to it at I. In that case, by tangent vs radius relation, $\angle BIO_R = 90^0$.

After extending $IO_R$ to cut the circle at $P$, we have $IP$ as the diameter with $\angle IXP = 90^0$.
Let the circle cuts $AC$ at $K$ such that $AXIK$ is cyclic.
For some reasons, angles marked with the same color are all equal. In particular, $\angle 2 = \angle 3$ is because of “angles in the alternate segment”.
Then, $\alpha = 90^0 – \angle green = \angle IXP – \angle green = \angle pink + \angle purple = \gamma$.
In the similar fashion, if we use the other given (i.e. $CY.AC = IC^2$), we have construct another circle and this will result in $\beta = \gamma$
From the fact that $\alpha , \beta , \gamma$ split the straight angle into three equal parts, we have $\beta = 60^0$
Result follows from $A = \beta$.
A: Well, here is a barycentric coordinates solution. 
Hopefully, someone else can find a nice synthetic solution.
Use the well-known identities $IB^2=\frac{ca(s-b)}{s}$ and $IC^2=\frac{ab(s-c)}{s}$.
We get that $BX=\frac{a(s-b)}{s}$ and $CY=\frac{a(s-c)}{s}$.
Now we calculate the coordinates of $X,Y$ as $$X(\frac{a(s-b)}{s} : c- \frac{a(s-b)}{s} : 0)$$ and $$Y(\frac{a(s-c)}{s} : b-\frac{a(s-c)}{s} : 0)$$
From the well known coordinate of $I(a:b:c)$, we just need to calculate the determinant, find a condition for the determinant to be $0$, and calculate $\angle A$.
We calculate the determinant of the matrix $\left[ \begin{array}{cc} a(s-b) & cs-a(s-b) & 0 \\ a(s-c) & 0 & bs-a(s-c) \\ a & b& c \end{array} \right]$.
Some brutal calculations show that the determinant is just $$\frac{1}{4}a(a-b-c)(a+b+c)(a^2-b^2+bc-c^2)$$
For this value to be $0$, we need $a^2=b^2-bc+c^2 = b^2+c^2-2bc \cos A$.
Therefore, we have $\cos A = \frac{1}{2}$, so $\angle A = \boxed{60^\circ}$
