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.

$$\begin{gathered} \therefore \frac{\mathrm{PT}}{\mathrm{TQ}}=\frac{\mathrm{AC}}{\mathrm{CB}} \\ \Rightarrow \frac{\mathrm{PT}}{\mathrm{TQ}}=1\left(\mathrm{AC}=\mathrm{CB}\right) \\ \Rightarrow \mathrm{PT}=\mathrm{TQ} \end{gathered}$$
In ∆CPT and ∆CQT,

seg PT ≅ seg TQ (Proved)

∠CTP = ∠CTQ (90º each)

seg CT ≅ seg CT (Common)

∴ ∆CPT ≅ ∆CQT (RHS congruence criterion)

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

Hence proved.