Question

In the given above figure, there are two concentric circles with centre 'O'. Chord AD of the bigger circle intersects the smaller circle at B and C. Show that AB = CD.

Answer 1

In two concentric circles with center' O'. $\overline{AD}$ is the chord of the bigger circle. $\overline{AD}$ intersects the smaller circle at B and C.

Construction: Draw $\overline{OE}$ perpendicular to $\overline{AD}$

Proof: AD is the chord of the bigger circle with center' O' and $\overline{OE}$ is perpendicular to $\overline{AD}$.

$\because \overline{OE}$ bisects $\overline{AD}$ (The perpendicular from the center of a circle to a chord bisect it)

$\therefore AE=ED$ ....(i)

$\overline{BC}$ is the chord of the smaller circle with center' O' and $\overline{OE}$ is perpendicular to AD.

$\because \overline{OE}$ bisects $\overline{BC}$ (from the same theorem)

$\therefore BE=CE$ .... (ii)
Subtracting the equation (ii) from (i), we get
$AE-BE=ED-EC$
$AB=CD$