Question

In the given figure O and O' are the centres of two intersecting circles and APB is parallel to OO' . Prove that AB = 2OO'.

Answer 1

A perpendicular from centre to a Chord bisects the Chord.

So , PN = NB &PM = AM------->(1)
as OO'NM is //gm
OO' = NM = PN + MP

AB = NB + PN + PM + AM
AB = 2(PN +MP) using eq. (1)
AB = 2OO'

Answer 2

From figure we can say that,
MP + PN = OO'. (1)

We know that a perpendicular drawn to any chord of the circle divides it into two equal parts,so

In circle O,
AM = MP. (2)

In circle O'
BN = NP. (3)

(2) + (3)
AM + BN = MP + NP
AM + BN = OO'. (4). [From (1)]

(1) + (4)
AM + MP + PN + NB = 2OO'
AB = 2OO'

Hope this helps... plsss make it the brainliest