Q. 54.1( 26 Votes )

In the figure 6.2

Answer :



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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