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  • The perpendicular bisectors of two chords of a circle intersect at its centre. (True/False)Option: 1 True<stron

The perpendicular bisectors of two chords of a circle intersect at its centre. (True/False)

Option: 1

True


Option: 2

False


Answers (1)

best_answer

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

We need to prove O' coincides with O.

Join OE and OF.

Now, E is the midpoint of chord AB

\\\Rightarrow \quad O E \perp A B \\\Rightarrow \quad \text{OE is the perpendicular bisector of AB}\\ \Rightarrow \quad \text{OE as well as O'E is the perpendicular bisector of AB} \\\Rightarrow \quad \text{O'E lies along OE}

Similarly, F is the midpoint of chord CD

\\\Rightarrow \quad O F \perp CD \\\Rightarrow \quad \text{OF is the perpendicular bisector of CD}\\ \Rightarrow \quad \text{OE as well as O'F is the perpendicular bisector of CD} \\\Rightarrow \quad \text{O'F lies along CD}

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

the point of intersection of O'E and O'F coincides with the point of intersection of OE and OF.

O' coincides with O.

Hence, the perpendicular bisectors of AB and CD intersect at O.

Posted by

Ritika Kankaria

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