Question

Prove that the perpendicular drawn from the center of a circle to a chord bisects the

chord.
​

Answer 1

Answer:

see below

Step-by-step explanation:

The perpendicular from the centre to the chord bisects the chord

In the diagram below, AB is the chord of a circle with centre O.

OM is the perpendicular from the centre to the chord.

Look at triangles OAM and OBM.

The hypotenuses (OA and OB) are the same, as they are both the radius of the circle.

OM is common to both triangles.

OMA and OMB are both right angles.

Triangles OAM and OBM are congruent (RHS), so it follows that AM = MB.

Therefore, M is the midpoint of AB, and the chord has been bisected.

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Take O as centre, chord AB and The perpendicular AD.

In triangle OAD and triangle OBD,

angle ODA = angle ODB (90 degree each)

OA = OB (Radius of circle)

OD = OD (Common)

Therefore, triangle OAD is congruent to triangle ODB (By RHS congruency rule)

AD = BD (CPCT)