Question

Fig. 3.8
P
10*. 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 API line PQ and seg BQ I line PQ.
Prove that, seg CP = seg CQ.​

Answer 1

Question

In a given figure

Seg AB is a diameter of a circle woth centre C. Line PQ is a tangent, which touches the circle at pt T. Seg AP perpendicular to line PQ and BQ perpendicular to line PQ.

Prove that seg CP= segCQ

ANSWER

since PQ is a tangent to the circle,

∠CTP = ∠CTQ = 90

Since ∠APT = ∠CTP = 90

AP || CT.

Similarly CT || BQ.

From association, we can say 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.

Hence PT = TQ.

Now in ΔCTP and ΔCTQ,CT is a common side, PT = TQ and ∠CTP = ∠CTQ = 90

So we can prove CP = CQ. (Pythagoras theorem or congruent triangle theorem)

Hope it helps you

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