In the figure 6.2

In the figure, we join PA and PD. Draw perpendiculars on AB and CD from P as PM and PN respectively.

∠AEP = ∠ DEP (given)

So, ∠PEN = ∠PEM (M and N are points on line AE and ED respectively)

In ΔPEN and ΔPEM,

∠PNE = ∠PME = 90°

∠PEN = ∠PEM

PE = PE (common)

Therefore, Δ PEN ≅ ΔPEM (by AAS congruence)

∴ PN = PM (by CPCT)

We know that The chords of a circle equidistant from the center of a circle are congruent.

So, AB = CD.

Hence proved.