In a triangle $ABC$, points $X$ and $Y$ are on $BC$ and $CA$ respectively such that $CX=CY$,$AX$ is not perpendicular to $BC$ and $BY$ is not perpendicular to $CA$.Let $\Gamma$ be the circle with $C$ as centre and $CX$ as its radius.Find the angles of triangle $ABC$ given that the orthocentres of triangles $AXB$ and $AYB$ lie on $\Gamma$.
1. **Lemma Analysis**: The lemma states that for a triangle \(ABC\) with sides \(a\) and \(b\) opposite to angles \(A\) and \(B\) respectively, if \(a \cos 2B = b \cos 2A\), then \(a = b\). This is based on the properties of the cosine function and the fact that it is decreasing in the interval \([0^\circ, 90^\circ]\).

2. **Orthocenters and Circle \(\Gamma\)**: Let \(H_1\) be the orthocenter of \(\triangle ABX\) and \(H_2\) be the orthocenter of \(\triangle AYB\). Given that \(CX = CY\), the circle \(\Gamma\) with center \(C\) and radius \(CX\) passes through both \(H_1\) and \(H_2\).

3. **Perpendiculars and Cyclic Quadrilateral**: Since \(AH_1 \perp BC\) and \(BH_2 \perp CA\), let \(BC \cap AH_1 = D\) and \(BH_2 \cap AC = Y\). Given \(CX = CH_1\) and \(AMXD\) is a cyclic quadrilateral (where \(M\) is the foot of the altitude from \(X\) on \(BC\)), we have \(\angle DXH_1 = 90^\circ - B\).

4. **Angle Calculation**: 
   \[
   \angle BCH_1 = 180^\circ - 2(90^\circ - B) = 2B
   \]

5. **Radius Calculation**: 
   \[
   r = H_1C = \frac{CD}{\cos 2B} = \frac{b \cos C}{\cos 2B}
   \]
   Similarly, in \(\triangle ABY\):
   \[
   r = \frac{a \cos C}{\cos 2A}
   \]

6. **Equating Radii**: 
   \[
   \frac{b \cos C}{\cos 2B} = \frac{a \cos C}{\cos 2A} \implies a \cos 2B = b \cos 2A
   \]

7. **Using the Lemma**: From the lemma, since \(a \cos 2B = b \cos 2A\), we conclude that \(a = b\).

8. **Conclusion**: Given \(a = b\) and the properties of the triangle, it follows that \(\triangle ABC\) must be a right-angled isosceles triangle.

\(\blacksquare\)

The final answer is \( \boxed{ \triangle ABC } \) is a right-angled isosceles triangle.