CAT Perpendicular from the Centre to a Chord - Practice Questions & MCQ

Theorem 3 : The perpendicular from the centre of a circle to a chord bisects the chord.

Let's prove this. Draw a circle on a tracing paper with center O and draw a chord AB. Fold the paper along a line through O so that a portion of the chord falls on the other. Let the crease cut AB at the point M. Then, ∠ OMA = ∠ OMB = 90° or OM is perpendicular to AB. You will observe that the point B coincide with A. Thus, MA = MB.

Also,

In the right triangles OMA and OMB

                           OA = OB         (Radii of a circle)

                          OM = OM         (common)

                    ∠ OMA = ∠ OMB          (90°)

Therefore,   ∆ AOB ≅ ∆ COD    ([by RHS-congruence)

Hence,             MA = MB

Converse of the above theorem

Theorem 4 : The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

M is the midpoint of the chord AB of a circle C(O,r)

In the triangles OMA and OMB

                           OA = OB         (Radii of a circle)

                          OM = OM         (common)

                          MA = MB          (given)

Therefore,   ∆ AOB ≅ ∆ COD    ([by SSS congruence rule)

This gives  ∠ OMA = ∠ OMB     (Corresponding parts of congruent triangles)

But,        ∠ OMA + ∠ OMB = 180°         (Linear Pair angle)

Therefore, ∠ OMA = ∠ OMB = 90°

Hence, OM ⊥ AB.