Show that the triangle formed by a chord and the centre of the circle is isosceles.
Answer: For a chord AB of a circle with centre O, the two sides OA and OB of ΔOAB are both radii, so OA = OB and ΔOAB is isosceles (and so ∠OAB = ∠OBA).
Step-by-step solution
Given: A circle with centre O and a chord AB
To find: Show that ΔOAB is isosceles
To find: Show that ΔOAB is isosceles
Idea: A triangle is isosceles if two of its sides are equal. The two sides that meet at the centre join the centre to points on the circle, so each is a radius.
- Let AB be a chord of a circle with centre O and radius r. Join OA and OB to form ΔOAB. (If AB is a diameter, O lies on AB and there is no triangle; so take AB not a diameter.)½ mark
- A and B lie on the circle, so OA = r and OB = r (radii of the circle).1 mark
- Two sides of ΔOAB are equal, so ΔOAB is isosceles, with base AB. As a result, its base angles are equal: ∠OAB = ∠OBA.½ mark
OA = OB because both are radii, so the triangle formed by a chord and the centre is always isosceles.
Answer to write in the exam
Let AB be a chord (not a diameter) of a circle with centre O. Join OA, OB.
OA = OB (radii of the same circle)
∴ ΔOAB is isosceles (two sides equal), and ∠OAB = ∠OBA (angles opposite equal sides).
Common mistakes that cost marks
- Saying the triangle is equilateral. AB equals the radius only in the special case ∠AOB = 60°.
- Forgetting the case of a diameter: then O lies on AB and no triangle is formed.
- Calling AB a radius. AB is the chord (the base); the equal sides are OA and OB.
How this can come in the exam
MCQ (1 mark)
AB is a chord of a circle with centre O and ∠OAB = 35°. Then ∠AOB is
- 35°
- 70°
- 110°
- 145°
Show answer
(C) 110°
OA = OB, so ∠OBA = ∠OAB = 35°. Then ∠AOB = 180° − 35° − 35° = 110°.
Try one yourself
PQ is a chord of a circle with centre O and ∠POQ = 100°. Find ∠OPQ.
Show answer
OP = OQ (radii), so ∠OPQ = ∠OQP = (180° − 100°) ÷ 2 = 40°.
More questions like this
- Show that if two such isosceles triangles (formed by a chord and the centre of the circle) have equal base length, they are congruent to each other.
- The line joining the centre of a circle and the midpoint of a chord of the circle is perpendicular to the chord.
- Can you explain why the converse of the result “the line joining the centre of a circle and the midpoint of a chord is perpendicular to the chord” is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord?
(Hint: Use the figure. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM.) - An isosceles triangle ABC is inscribed in a circle, with AB = AC. Show that the altitude from A to BC passes through the centre of the circle.
- Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.