Question

In the following figure, the line ABCD is perpendicular to PQ ; where P and Q are the centres of the circles. Show that:
i) AB = CD
ii) AC = BD

Answer 1

Given :-$PQ$ is perpendicular to $AB$

$\left(i\right)AB=CD$

$\left(ii\right)AC=BD$

$\left(i\right)$ consider triangle $PBC$

$BC$ is chord of small circle

and $P$ is center.

$\Rightarrow PO$ is perpendicular to $BC$ (Given)

$\Rightarrow BO=CO$...(1) [perpendicular from center to chord divides the chord equally]

consider triangle $QAD$

$Q$ is center of big circle

$AD$ is chord.

and $QO\perp AD$ (Given)

$\Rightarrow AO=OD$ [perpendicular from center to a chord divide the chord equally. ]

$\Rightarrow (AB+BO)=(OC+CD)$

$\Rightarrow (AB+BO)=(BO+CD)$ From $\left(1\right)BO=CD$

$\left(ii\right)$ we know

$AB=CD$(from part(i))

$AB+BC=CD+BC$ (added $BC$ on both sides)

$AC=BD$