ABC and DBC are two equilateral triangles on the same base BC. A point P is taken on the circle with centre D, radius BD. Show that PA, PB, PC are the sides of a right triangle.

Letting $$\angle DBP=\theta$$ and using the law of sines in the relevant triangles, PB and PC can be expressed in terms of the radius r and $$\theta$$, specifically $$PB=2r\cos\theta$$ and $$PC = r(\sqrt{3}\cos\theta+\sin\theta)$$. Computing $$AP^2$$ through the geometry of the equilateral triangle configuration gives $$AP^2 = r^2(1+6\cos^2\theta+2\sqrt{3}\sin\theta\cos\theta)$$, and directly computing $$PB^2+PC^2$$ gives the identical expression. Since $$AP^2 = PB^2+PC^2$$, triangle with sides PA, PB, PC is right-angled at the vertex opposite PA.