Question

In the figure, 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 1 line
PQ and seg BQ I line PQ.
Prove that, seg CP seg CQ.​

Answer 1

Answer:

Seg AB is a diameter of a circle with centre C

∴AC=CB (Radii of the circle)

Join CP,CT and CQ

It is given that line PQ is a tangent, which touches the circle at point T.

∴∠CTP=∠CTQ=90

o

(Tangent at any point of a circle is perpendicular to the radius through the point of contact)

⇒ seg CT⊥ line PQ

Also, seg AP⊥ line PQ and seg BQ ⊥ line PQ.

∴ seg AP seg CT || seg BQ

We know, the ratio of the intercepts made on a tranversal by three parallel lines is equal to the ratio of the corresponding intercepts made on any other transversal by the same parallel lines.

∴

TQ

PT

=

CB

AC

⇒

TQ

PT

=1(AC=CB)

⇒PT=TQ

In △CPT and △CQT,

seg PT≅ seg TQ (Proved)

∠CTP=∠CTQ (90

o

each)

seg CT≅ seg CT (Common)

∴△CPT≅△CQT (RHS congruence criterion)

⇒ seg CP≅ seg CQ(Corresponding parts of congruent triangles)

Hence proved