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Explain the theorem:- Chords equidistant from the centre of a circle are equal in length?
Asked by | 27 Feb, 2009, 05:31: PM

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.

Now,

OM=ON(as it's given that the chords are equidistant  from the center)

So,

triangle OMB congruent to triangle OND (RHS rule)

MB=ND( C.P.C.T.)

We know that the perpendicular from the centre of the circle bisects the chord.

So

MB=1/2(AB)

and

ND=1/2(CD)

Thus, MB=ND implies

1/2(AB)=1/2(CD)

i.e.

AB=CD

Hence proved.

Answered by | 28 Feb, 2009, 09:07: AM
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