Question

help this question fast....​

Answer 1

Answer:

Step-by-step explanation:

Given circles having centres O and O’ intersect at P.

Also CD ||OO’.

To prove CD = 2OO’

Construction Draw OA and O’B perpendicular to CD from O and O’, respectively.

Proof OA⊥CD

∴OA bisects the chord.

Also, CP is perpendicular from the centre to the chord bisects the chord.

$AP=\frac{1}{2}CP$

$CP=2AP$.......(i)

Similarly, O'B ⊥ PD

$BP=\frac{1}{2} PD$

$PD=2BP$.......(ii)

$now, CD=CP+DP=2AP+2BP$ [ from equation (i) and (ii) ] .......(iii)

in quadrilateral ABO'O

OA = O'B [ two lines perpendicular to same line ]

ABO'O is a parallelogram

AB = OO' [opposite sides of a parallelogram are same ]

CD = 2AB = 200' hence proved