Question

$AB$ and $CD$ are two parallel chords of a circle (radius$=r$) and a line $\ell$ is the perpendicular bisector of $AB$.
Then which of the following is true?

• A. $AB=CD$
• B. $\ell$ is the perpendicular to $CD$
• C. $\ell$ is the perpendicular bisector of $CD$
• D. None of the above

Answer 1

The correct option is C $\ell$ is the perpendicular bisector of $CD$

$AB$ and $CD$ are two parallel chords of a circle.

$l$ is a line which perpendicularly bisects $AB$ at $M$.

To find out- which of the given options is true.

Solution-

The line $l$ perpendicularly bisects $AB$ at $M$.

$\therefore$ $l$ is the diameter of the given circle.

A line, bisecting a chord of a circle perpendicularly, is the radius i.e. diameter of the circle passing through the point of right bisection.

Let $l$ meet $CD$ at $N$.

We bisect $l$.

The mid point is $O$.

$\therefore O$ is the centre of the given circle and also we join $OC$ and $OD$ which are the radii\\ of the given circle.

So, $OM\perp AB$

$\Rightarrow \angle AMN={90}^o=\angle BMN$

Again $AB\parallel CD$

$\therefore \angle AMN={90}^o=\angle OND$ and $\angle BNM={90}^o=\angle ONC$ ......(alternate angles)

So, between $\Delta OCN$ and $\Delta OND$, we have

$OC=OD$ ......(radii of the same circle)

$ON$ is the common side,

$\angle AMN={90}^o=\angle BMN$

$\therefore$ by SAS test,

$\Delta OCN\cong \Delta OND$

$\therefore CN=DN$

$\therefore$ $l$ is the perpendicular bisector of $CD$.