Question

Prove that, if a diameter of
a circle bisects two chords
of the circle then those two
chords are parallel to each
other.​

Answer 1

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Let AB and CD be two chords of a circle whose centre is O, and let PQ be a diameter bisecting chords AB and CD at L and M respectively. Since PQ is a diameter. So, it passes through the centre O of the circle.

Now,

L is mid-point of AB.

OL⊥AB [ Line joining the centre of a

circle to the mid-point of a chord is perpendicular to the chord]

∠ALO=90

o

Similarly, ∠CMO=90

o

Therefore, ∠ALO=∠CMO

But, these are corresponding angles.

So, AB∥CD.

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Answer 2

Given :

• AB and CD are chords

• Centre of the circle is O

• PQ is diameter of the circle which intersects the two chords in 90° angle.

To Prove :

• AB || CD

Solution :

We know that When chord intersect diameter so they bisect and form a 90° Angle

it means it divides the chord equally

∠AMN = CNM

= 90°

PQ is transversal which is passing through the lines AB and CD

Therefore, AB || CD