Question

The figure shows two concentric circles and $AD$ is a chord of larger circle.
Prove that: $AB=CD$

Answer 1

Drop $OP\perp AD$
$\therefore OP$ bisects $AD$
(perpendicular drawn from the centre of a circle to a chord bisects it)
$\Rightarrow AP=PD...\left(i\right)$
Now, $BC$ is a chord for the inner circle and $OP\perp BC$
$\therefore OP$ bisects $BC$
(perpendicular drawn from the centre of a circle to a chord bisects it)
$\Rightarrow BP=PC...\left(ii\right)$
Subtracting $\left(i\right)$ and $\left(ii\right)$
$AP-BP=PD-PC$
$\Rightarrow AB=CD$