Question

Prove that the line drawn from the centre of a circle to bisect a chord is perpendicular to the chord.

Answer 1

Hint: Consider the triangle OCA and OCB. Try proving that the triangles are congruent. Observe that OA and OB are radii of the circle and hence area equal, AC and BC are equal since C is the midpoint and OC is the common side of these triangles. Hence prove that the triangles OAC and OBC are congruent and hence prove that the angles OCA and OCB are equal. Finally, use the fact that the angles OCA and OCB add up to 180 since they form a linear pair. Hence prove that both the angles OCA and OCB are right angles and hence prove that OC is perpendicular to AB.

Complete step-by-step answer:

Given: AB is a chord of the circle, and O is the centre of the circle. OC bisects AB.
To prove: OC is perpendicular to AB.
Construction: Join OA and OB.
Proof:
In triangles OAC and OBC, we have
OA = OC(radii of the same circle)
AC = BC (OC bisects AB)
OC = OC(common side)
Hence, we have $\Delta OAC\cong \Delta OBC$(By SSS congruence criterion)
Hence, we have $\angle OCA=\angle OCB$(Corresponding parts of congruent triangles)
Now, we have $\angle OCA+\angle OCB=180{}^\circ $(Linear pair axiom)
Hence, we have $2\angle OCA=180{}^\circ \Rightarrow \angle OCA=90{}^\circ $
Hence OC is perpendicular to AB.
Hence proved.

Note: [1] This result is a direct consequence of the fact that the median from the vertex of the isosceles triangle opposite to the unequal side(often referred to as the base) is an altitude, an angle bisector and a perpendicular bisector also.
Since OA = OB, AOB is an isosceles triangle, and OC is the median from vertex O. Hence OC is also an altitude, and hence OC is perpendicular to AB.