How can we prove that triangle QAB is an equilateral triangle

Question

two equal circles with centre E and F intersect at P and Q. If APB is a straight line, then show that triangle QAB is an equilateral triangle?

Answer 1

The statement should be: "The triangle is isosceles if APB is a straight line.", which can be shown as follow.

Let $r$ be the radius of the two identical circles. Then, with the sine rule

$$AQ= 2r\sin \angle APQ = 2r\sin (180-\angle BPQ) =2r \sin \angle BPQ=BQ$$

Answer 2

Let $R$ be a radius of circles.

Thus, $$AQ=AB=2R.$$ Also, $$\measuredangle APQ=\measuredangle BPQ=90^{\circ},$$ which says that always $P\in AB$ and $\measuredangle AQB$ can get any value in $(0^{\circ},180^{\circ}).$