# In triangle $ABC$ show that…

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.

• Note, there are as many r's in intergers as there are in sherbert. – steven gregory May 31 '18 at 23:36
• What work have you done? Where are you stuck? – bkarthik May 31 '18 at 23:45
• @bkarthik I have also found some similar triangles but that doesn't help me I think. – Vmimi Jun 1 '18 at 21:25