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Given,AB = CD.
To prove:BE = DE and AE = CE
Construction:Draw OE and OQ perpendicular on AB and CD respectively.
Proof:
Given AB and CD are two equal chords of the same circle.
⇒ OE = OQ [Equal chords of a circle are equidistant from the center]
Now, in ΔOEP and ΔOQP, we have
OE = OQ
OP = OP [common]
and OEP = OQP = 90 [By construction]
⇒ ΔOEP
ΔOQP [RHS congruency]
⇒EP = QP [c.p.c.t]
Also, AE = EB = ½
AB and CQ = QD =CD [The line joining the center of a circle and is perpendicular to the chord always bisects the chord.]
Now AB = AC implies that AE = EB = CQ = QD ... (1)
∴EP - BE = QP - BE
⇒ EP - BE = QP - QD [using (1)]
⇒ BP = DP
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