Assume $ \triangle ABC$ is a right triangle where $\angle C= 90°$. Suppose that $ \overline {AX} $ bisects $\angle A$ and point $X$ lies on $BC$. Assume that the circumcircle of triangle $AXB$ intersects $AC$ on $Y$.
a) Show that if lengths $BC$ and $CY$ are intergers divisible by a prime $p$, then $AY$ is also an intenger divisible by $P$.
b) Show that if $CY= k$ and $BC= 3k$ ($k$ is an interger), then the lengths of sides of the triangle $ABC$ must be intergers.
This is what I have done:
Applying power of point of $C$, we get:
$$CY (CA) = (CX)(CB)$$
$$CY(CY+AY)= CB(CB -XB)$$
$$ (CY)^2+ CY(AY) = (CB)^2 - CB(XB)$$
$$CY(AY) = (CB)^2 - (CY)^2 - CB(XB)$$
$$ AY = \frac{(CB)^2 - (CY)^2 - CB(XB)}{CY}$$
But this doesn't prove that $AY$ is divisible by $p$ or that it is an interger.