Question

Q)3:- A perpendicular drawn from the centre of a circle on its choud bisects the chord.
(slave the theovem).​

Answer 1

Answer:

the perpendicular from the center of a circle to a chord bisects the chord. So, C'C bisects AB. Since CC' ⊥ AB and line segment joining the centers of two circles bisects the common chord. Therefore, C' is the mid-point of AB

Answer 2

Theorem :

★ The Perpendicular to a chord drawn from the centre of the Circle bisects the chord .

Given :

Chord AB

Circle with centre O

OM perpendicular to AB

To Prove :

AM = Bm

Construction :

Join OA and OB

Proof :

From ∆ AMO and ∆ BMO

OA = OB ( Radius of same circle )

/_ AMO = /_ BMO ( Each = 90° ) [ /_ = Angle ]

OM = OM ( Common )

=> ∆ AMO is congruent to ∆ BMO ( RHS )

=> AM = BM ( cpct )

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