If the circles shown in the below diagram have the centers A and B, and they have the same radius then ΔAPQ≅ΔBPQ.

The statement is true.

If the circles have the same radius, the problem is solved. As a result, the two circles are similar.
In ∆APQ and ∆BPQ
AP=BP in triangles (Radius of the congruent circles)
AQ = BQ (Radius of the congruent circles)
PQ = PQ (Common side)
By using the S.S.S congruence rule, APQ≅ΔBPQ