Question

Prove that the line joining the midpoint of a chord to the centre of a circle is perpendicular to the chord.

Answer 1

We know that $AB$ is a chord of the circle and $C$ is the midpoint of it.

so in $△ACOand△BCO$

$AC=BC$ (as $C$ is mid point of $AB$)

$OC=OC$ (common side)

$OA=OB$ (radius of the circle)

$\therefore$ By SSS congruency condition $△ACO\cong △BCO$

Now, as we know that Corresponding Parts of Congruent triangles are equal

$\therefore \angle OCA=\angle OCB$

But$\angle OCA+\angle OCB=180°$ ($\because$ they make a linear pair )

$\therefore 2\times \angle OCA=180°$

$\Rightarrow \angle OCA=\angle OCB=90°$

$\Rightarrow$OC $\perp$ AB.

Hence proved.