Question

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Answer 1

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Given :-

OO' Parallel to CD  OR we can say that OO' Parallel to PD + PC ( CD = PD + PC ) ------->>> X

To Prove :-

CD = 2OO'

Construction :-

Draw perpendicular from O' to PD and form O to PC and let the meeting point of both be A and B respectively

Solution :-

PD and PC are chord of the circle with center O' and O respectively

O'A is perpendicular bisector of PD and OB is perpendicular bisector of PC ( Perpendicular line from center of the chord bisect the chord )

Therefore , ∠OBP = ∠O'BP = 90° -------->> Y

PA = AD

PB = BC

From (x) and( y) Quadrilateral ABOO' is a rectangle

So ,

O'A = OB

AB = OO' ( or , PA + PB = OO' )

We can see that ,

$CD=PD+PC=PA+AD+PB+BC=2PA+2PB=2(PA+PB)$

As  , PA = AD and PB = BC

Since , OO' = PB + PA

CD = 2(PA +PB )

Hence , CD = 2OO'

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