In figure 3.90, seg AB is a diameter of a circle with centre C. Line PQ is a tangent, which touches the circle at point T. seg AP line PQ and seg BQ line PQ. Prove that, seg CP ≅seg CQ.

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.