Question

The perpendicular bisectors of two chords of a circle intersect at its centre . Prove it​

Answer 1

Answer:

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Step-by-step explanation:

Perpendicular bisectors of two chords of a circle intersect at centre. Two parallel chords of a circle, which are on the same side of centre,subtend angles of 72∘and144∘ respetively at the centre Prove that the perpendicular distance between the chords is half the radius of the circle.

Answer 2

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Question:-

★the perpendicular bisector of two chords of a circle intersect at its centre. Prove it.

Given:-

★ AB and CD are two chords of a circle C(O,r) and let the perpendicular bisectors O'F of AB and CD respectively meet at O'

To prove:-

★ O' coincides with O.

Construction:-

★ Join OE and OF.

Proof:-

★E is the midpoint of chord AB

==> OE ⊥ AB

==> OE is the perpendicular bisector of AB

==> OE as well as O'E is the perpendicular bisector of AB

==> O'E lies along OE

Similarly, F is the midpoint of chord CD

==> OF⊥ CD

==> OF is the perpendicular bisector of CD

==> OF as well as O'F is the perpendicular bisector of CD

==> O'F lies along OF

Thus, O'E lies along OE and O'F lies along OF

==> The point of intersection of O'E and O'F coincides with the point of intersection of O'E and OF

==> O' coincides with O

Hence, the perpendicular bisector of ab and CD intersects at O

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