To Prove: seg CP ≅seg CQ
Construction: Join CP, CQ and CT
Figure:

Since PQ is a tangent to the circle, ∠CTP = ∠CTQ = 90
Since ∠APT = ∠CTP = 90
AP || CT.
Similarly,
CT || BQ.
So, we can say that,
AP || CT || BQ
AB is a line cutting all three parallel lines.
AC = CB (Radius of the circle, and AB is diameter, C is center)
Since C is center point of line AB cutting parallel lines.
We can say these parallel lines are equal distance.
Therefore, PT = TQ.
Now in ΔCTP and ΔCTQ,
CT is a common side, PT = TQ and ∠CTP = ∠CTQ = 90
CP = CQ. (Pythagoras theorem or congruent triangle theorem)
Hence, Proved.