# 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.

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