Question

prove that chords equidistant from the centre of a circle are equal in length

Answer 1

Continue from the above picture..

So, AN = DM
1/2 AB = 1/2 CD
Therefore, AB = CD

Hence proved, the chords equidistant from the centre of the circle are equal in length.

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Answer 2

From theorem 6.7
Chords equidistant from the Centre of the circle are equal distance.

Given:C is the Centre with O. AB and CD are two chords of the circle where OX is perpendicular to AB and OY is perpendicular to CD and OX=OY

To prove:-AB=CD
Proof::- in tri(AOX) and tri(DOY)
Angle OXA = angle OYD (both 90° given)
OA=OD (radius)
OX=OY (Given)
. SO,..
tri(AOX) =tri(DOY) ( by RHS)
AX=DY..........(1) (by C. P. C. T)

For CHORD AB
OX is perpendicular to AB
X bisects AB
AB=2AX...........(2)

FOR CHORD CD
OY is perpendicular to CD
Y bisects CD
CD=2DY............. (3)

From 1
AX=DY
2AY=2AY
Hence, AB=CD (from 2&3)



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