In the adjoining figure, P is a point of intersection of two circles with centres C and D. If the straight line APB is parallel to CD, prove that AB = 2CD.

From C draw CL perpendicular to AB and from D drawn DM perpendicular to AB.

From figure,

LCDM is a rectangle.

∴ ML = CD (Opposite sides of rectangle are equal).

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

LP = $\dfrac{1}{2}$ AP and MP = $\dfrac{1}{2}$ PB

From figure,

LM = LP + PM

$\Rightarrow LM = \dfrac{1}{2} AP + \dfrac{1}{2} PB \\[1em] \Rightarrow LM = \dfrac{1}{2} (AP + PB) \\[1em] \Rightarrow LM = \dfrac{1}{2} AB \\[1em] \therefore CD = \dfrac{1}{2} AB \\[1em] \Rightarrow AB = 2CD.$

Hence, proved that AB = 2CD.