Question

In the given 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 ⊥ line PQ and seg BQ ⊥ line PQ. Prove that, seg CP ≅ seg CQ.

Answer 1

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 = 90º     ...(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.

\[\therefore \frac{PT}{TQ} = \frac{AC}{CB}\]

\[ \Rightarrow \frac{PT}{TQ} = 1 \left( AC = CB \right)\]

\[ \Rightarrow PT = TQ\]   ...(1)

In ∆CPT and ∆CQT,

seg PT ≅ seg TQ    ...[From (1)]

∠CTP = ∠CTQ    ...(Tangent theorem)

seg CT ≅ seg CT   ...(Common)

∴ ∆CPT ≅ ∆CQT     ...(SAS congruence criterion)

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

Hence proved.