केंद्र से जीवा पर लंब

In Mathematics, a chord is a line segment that joins two points on the circumference of a circle. We know that the longest chord of a circle is a diameter that passes through the centre of a circle. In this article, the theorem related to the perpendicular from the centre to a circle and its proof, and the converse of this theorem in detail.

Perpendicular from the Centre to a Chord – Theorem and Proof

Theorem:

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

Proof:

Fig. 1

Consider a circle with centre $O$ shown in Fig. 1

$AB$ is a chord such that the line $OX$ is perpendicular to the chord $AB$. ($OX\perp AB$)

We need to prove: $AX=BX$

Consider two triangles $OAX$ and $OBX$

$\angle OXA=\angle OXB=90^{\circ }$

$OX=OX$ (Common side)

$OA=OB$ (Radii)

By using the RHS rule, we can prove that the triangle $OAX$ is congruent to $OBX$.

Therefore,

$\triangle OAX\cong \triangle OBX$

Hence, we can say that $AX=BX$ ( Using CPCT)

Thus, the perpendicular from the centre of a circle to a chord bisects the chord, is proved.

The Converse of this Theorem:

The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord

Proof:

Consider the Fig. 1

Assume that $AB$ is the chord of a circle with centre $O$.

The centre $O$ is joined to the midpoint $X$ of the chord $AB$.

Now, we need to prove $OX\perp AB$

Join $OA$ and $OB$ and the two triangles formed are $OAX$ and $OBX$.

Here,

$OA=OB$ (Radii)

$OX=OX$ (Common side)

$AX=BX$ (As, $X$ is the midpoint of AB)

Therefore, we can say that $\triangle OAX\cong \triangle OBX$.

Thus, by using the RHS rule, we get

$\angle OXA=\angle OXB=90^{\circ }$

This proves that the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.Hence, the converse of this theorem is proved.