Let $$ABC$$ be a triangle with $$AB = AC$$. If $$D$$ is the midpoint of $$BC, E$$ is the foot of the perpendicular drawn from $$D$$ to $$AC$$ and $$F$$ the mid-point of $$DE$$, prove that $$AF$$ is perpendicular to $$BE$$.

The line x + 3y = 0 is a diameter of the circle $${x^2} + {y^2} - 6x + 2y = 0\,$$.

The general solutions of $$\,\sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x$$ is

If the two circles $${(x - 1)^2} + {(y - 3)^2} = {r^2}$$ and $${x^2} + {y^2} - 8x + 2y + 8 = 0$$ intersect in two distinct points, then