Question

A straight line is drawn cutting two equal circles as shown in the figure and passing through the mid-point $M$ of the line joining their centers $O$ and $O^{\prime }$. Prove that the chords $AB$ and $CD$, which are intercepted by the two circles, are equal.

Answer 1

Given: A straight line $AD$ intersects two circles of equal radius at $A,B,C$ and $D$.
The line joining the centers $O{O}^{{}^{\prime }}$ intersect $AD$ at $M$ and $M$ is the midpoint of $O{O}^{{}^{\prime }}$.

To prove : $AB=CD$,
Construction : From $O,$ draw $OP\perp AB$ and from $O^{{}^{\prime }}$, draw $O^{{}^{\prime }}Q\perp CD$

Proof:
In $△OMP$ and $△O^{{}^{\prime }}MQ$,
$\angle OMP=\angle O^{{}^{\prime }}MQ$ (vertically opposite angles) and also $\angle OPM=\angle O^{{}^{\prime }}QM(each={90}^o)$
$OM=O^{{}^{\prime }}M$ (Given)

By Angle-Angle-Side criterion of congruence,
$\therefore △OMP\cong △O^{{}^{\prime }}MQ$ (by $AAS$)
The corresponding parts of the congruent triangles are congruent
$\therefore OP=O^{{}^{\prime }}Q(c.p.c.t)$

Also, we know that two chords of a circle or of equal circles which are equidistant from the center are equal.
$\therefore AB=CD$