Show that the triangle formed by a chord and the centre of the circle is isosceles.

Question

Homi · Accepted Answer

Understand the Question

A chord connects two points on the circumference of a circle.
Joining the centre of the circle to both endpoints of the chord forms a triangle.
The line segments connecting the centre to any point on the circle are radii, so two sides of the triangle are equal in length ( $OA = OB$ ).
A triangle with at least two equal sides is an isosceles triangle.

Step 1 · Construct the Triangle

Consider a circle with centre $O$ and a chord $AB$ .

Join $OA$ and $OB$ to form $\Delta OAB$ . Diagram 1

Step 2 · Prove the Triangle is Isosceles

In $\Delta OAB$

$OA$ and $OB$ are radii of the same circle.

Since all radii of a circle are equal $OA = OB$

Since two sides of $\Delta OAB$ are equal, $\Delta OAB$ is an isosceles triangle.

Answer

Hence proved, the triangle formed by a chord and the centre of the circle is an isosceles triangle ( $OA = OB$ ).

Common Mistakes

Assuming an Equilateral Triangle: Assuming the chord length is also equal to the radius. The chord $AB$ equals the radius only when the central angle is $60^\circ$ ; in general, only $OA = OB$ , making it strictly an isosceles triangle.
Diameter Case: If the chord passes through the centre (a diameter), $O$ lies on the chord $AB$ , resulting in a straight line segment rather than a triangle.

Question 1