Asked by pcoolboy | 28th Feb, 2010, 03:10: PM
assume that two equal chords AB,CD of a circle intersect at P.
Let O be the center of the circle.
drop OM,ON perpendiculars from O to AB,CD respectively.
since perpendicular from the center bisects a chord,
so Mand N become the mid points of AB,CD respectively.
consider right triangles OPN,OPM
OP=OP common side
OM=ON... equal chords are equidistant from the center
so OPN cong to OPM... RHS rule.
NC=AM=DN=MB.. since the chords are equal, their half parts also are equal.
but PM=PN, proved above,
Answered by | 28th Feb, 2010, 05:28: PM
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