if two equal chords of a circle intersect within the circle prove that the segments of the one chord are equal to the corresponding segments of the Other chord
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Given : Let AB and CD be two equal chords of a circle having centre O intersecting each other at point E within the circle.
To Prove :-
(i) AE = CE
(ii) BE = DE
Construction : Draw OM perpendicular at AB, ON perpendicular at CD. Join OE.
Proof :- in rt. angle d ∆s OME and ONE,
angle OME = angle ONE [Each = 90°]
OM = ON
[ because Equal chords are equidistant from the center]
hyp. OE = hyp. OE [Common]
Therefore, By RHS Congruence,
∆OME and ∆ONE are congruent
Therefore, ME = NE ....(1)
Now; O is the centre of circle and OM is perpendicular at AB
Therefore, AM = 1/2 of AB ...(2)
[Because, Perpendicular from the centre bisects the chord]
Similarly, NC = 1/2 of CD ....(3)
But AB = CD [Given]
From (2) and (3),
AM = NC ....(4)
Also, MB = DN .....(5)
Adding (1) and (4),
AM + ME = NC + NE
Hence, AE = CE.
Now, AB = CD (Given)
AE = CE (Proved)
Subtracting AB - AE = CD - CE
Hence, BE = DE.
