In the diagram, two circles with centres O and P respectively intersects at A. The two circles have the same radius and AB // OP. Show that AB = OP.

1. Identify equal radii and isosceles triangles

Since the two circles have the same radius,
$OA=PA=PB.$
Hence $\triangle OAP$ and $\triangle APB$ are isosceles triangles.

2. Use parallel lines and base angles to obtain equal angles

Since $AB \parallel OP$,
$\angle PAB = \angle APO.$
Also, since $OA=PA$, in $\triangle OAP$,
$\angle AOP=\angle APO.$
Since $PA=PB$, in $\triangle APB$,
$\angle PAB=\angle ABP.$
Therefore,
$\angle ABP=\angle AOP.$

3. Prove congruence and conclude the required equality

In $\triangle APB$ and $\triangle PAO$,
$\angle PAB=\angle APO,$
$\angle APB=\angle PAO,$
and
$AP=PA.$
Thus $\triangle APB \cong \triangle PAO$ by ASA.

Therefore, corresponding sides are equal:
$AB=OP.$