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.

Answer: CT ⊥ PQ and T is the midpoint of PQ, so CT is the perpendicular bisector of PQ and CP = CQ.

Draw seg CT. CT ⊥ line PQ (radius ⊥ tangent). Also AP ⊥ PQ and BQ ⊥ PQ, so AP ∥ CT ∥ BQ.

Three parallel lines make proportional intercepts on two transversals AB and PQ:

(AC = CB, radii). So T is the midpoint of PQ.

In \(\triangle CTP\) and \(\triangle CTQ\): PT = QT, \(\angle CTP = \angle CTQ = 90^\circ\), CT is common. So \(\triangle CTP \cong \triangle CTQ\) (SAS test).

\(\therefore\) seg CP ≅ seg CQ (c.s.c.t.).