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In a cyclic quadrilateral ABCD, let the diagonals AC and BD intersect at X. Let the circumcircles of triangles AXD and BXC intersect again at Y. If X is the incentre of triangle ABY, show that ∠CAD = 90°.

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Given that X is the incentre of triangle ABY,

we have ∠BAX = ∠XAY.

Therefore, ∠BDC = ∠BAC = ∠BAX = ∠XAY = ∠XDY = ∠BDY.

This shows that C, D, Y are collinear.

Therefore, ∠CYX  + ∠XYD = 180°.

But the left-hand side equals (180° - ∠CBD) + (180° - ∠CAD).

Since ∠CBD = ∠CAD,

we obtain 180° = 360° - 2∠CAD.

This shows that ∠CAD = 90°

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