$AB$ is the common chord of two circles of equal radius. If a line through $B$ meets the two circles at $P$ and $Q$, then $\triangle PAQ$ is—

Because the two circles have equal radii, the common chord $AB$ subtends equal angles in each, making $\angle APB = \angle AQB$. Equal base angles in $\triangle PAQ$ force $AP=$, so the triangle is isosceles.