In figure 3.97, circles with centres C and D touch internally at point E. D lies on the inner circle. Chord EB of the outer circle intersects inner circle at point A. Prove that, seg EA ≅ seg AB.

Answer: ED is a diameter of the inner circle, so \(\angle EAD = 90^\circ\); the perpendicular from centre D bisects chord EB.

By the theorem of touching circles, E, C, D are collinear. D lies on the inner circle and C is its centre, so seg ED is a diameter of the inner circle.

Draw seg AD. \(\angle EAD = 90^\circ\) (angle inscribed in a semicircle), so seg DA ⊥ chord EB.

D is the centre of the outer circle and EB is its chord. The perpendicular drawn from the centre to a chord bisects the chord.

\(\therefore\) seg EA ≅ seg AB.