Explain the theorem:- Chords equidistant from the centre of a circle are equal in length?
Rememebr that the distance of a point from a line or line segment is always the perpendicular distance.
Consider that two chords AB and CD of a circle with center O.
Let the perpendicular from O meet AB at M and to CD at N.
Consider triangles OMB and OND ,
angle OMB = angle OND(90 degrees each)
So we see that both triangles are right triangles.
OM=ON(as it's given that the chords are equidistant from the center)
triangle OMB congruent to triangle OND (RHS rule)
We know that the perpendicular from the centre of the circle bisects the chord.
Thus, MB=ND implies
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