Help with Circle theorem question

I am writing notes for my exam and I would like to bring this question in as an example but I cant remember how to solve it!

Two circles have a common chord AB. P is a point on the line through A and B tangents PQ and PR are drawn prove that PQ = PR.

Please help me

You did not define PQ and PR well. In context, I would suppose that PQ is tangent to one circle and PR is tangent to the other, and Q and R are the respective tangent points.

If that is the case, then the line PAB intersects both circles at A and B. The power of P with respect to either circle is (PA)(PB). If P is exterior to the circles, then the power is the square of the tangent distance.

PQ² = PR² = (PA)(PB)
PQ = PR

I believe for this to be true, the radiuses of the circles have to be the same. I could be wrong about that, but I'm going to assume they are.

Say C is the center of the circle containing Q, and D the center of the other circle.

then CQ = DR, by definition of the radius.
Angle Q = Angle R = 90 degrees, so if PC=PD, then PCQ = PDR, by the HL-theorem for right triangles.

Now we want to show PC=PD.

Let E be the intersection of AB and CD. Then EC = ED, and all angles at E are right.
I'm not entirely sure why this is true.
(edit: It's true because since CA, DA, CB, and DB are radii, ACBD is a rhombus, and the diagonals of a rhombus have this property)

But then EC=ED and E=E and EP=EP, so by SAS we have that ECP = EDP.

This proves that PC=PD, which proves PCQ = PDR, which proves PQ=PR.