Prove that the perpendicular bisector of a chord passes through the centre of the circle.

Question

Homi · Accepted Answer

Understand the Question

A circle's centre is equidistant from any two points on its circumference because all radii are equal ( $OA = OB$ ).
To prove that the perpendicular bisector of a chord passes through the centre, we join the centre $O$ to the chord's midpoint $M$ and prove that $OM$ is perpendicular to chord $AB$ using triangle congruence ( $\text{SSS}$ criterion).
Since there is only one unique perpendicular to $AB$ at midpoint $M$ , the perpendicular bisector must be the line containing $OM$ , which passes through $O$ .

Step 1 · Set up and Prove Triangle Congruence

Let $O$ be the centre of the circle and $AB$ be a chord.

Let $M$ be the midpoint of $AB$ , so: $AM = MB$

Join $OA$ , $OB$ , and $OM$ .In $\Delta OAM$ and $\Delta OBM$ :

By $\text{SSS}$ congruence criterion: $\Delta OAM \cong \Delta OBM$

Step 2 · Show $OM \perp AB$ and Conclude

Since corresponding parts of congruent triangles are equal (CPCT): $\angle OMA = \angle OMB$

Since $\angle OMA$ and $\angle OMB$ form a linear pair on straight line $AB$ : $\angle OMA + \angle OMB = 180^\circ$

\begin{aligned} 2 \angle OMA &= 180^\circ \\[0.6em] \angle OMA &= \dfrac{180^\circ}{2} \\[0.6em] &= 90^\circ \end{aligned}

$\angle OMA = \angle OMB = 90^\circ \implies OM \perp AB$

Since $OM$ is perpendicular to $AB$ at its midpoint $M$ , $OM$ is the unique perpendicular bisector of chord $AB$ .

Therefore, the perpendicular bisector of chord $AB$ passes through the centre $O$ .

Answer

Hence proved, the perpendicular bisector of a chord passes through the centre of the circle.

Common Mistakes

Assuming $OM \perp AB$ without Proof: Assuming $\angle OMA = 90^\circ$ directly instead of using triangle congruence $\Delta OAM \cong \Delta OBM$ and linear pair property.
Uniqueness of Perpendicular: Forgetting to mention that at a given point on a line, only one perpendicular can be drawn, ensuring $OM$ is indeed the perpendicular bisector.

Question 5