Chords of a circle that subtend equal angles at the centre are equal.
Step-by-step solution
To find: Show that AB = ED
Idea: This is the converse of “equal chords subtend equal angles at the centre”. The equal angles at the centre lie between two pairs of equal radii, which is exactly the SAS pattern.
- In ΔACB and ΔDCE: AC = DC (radii) and BC = EC (radii).1 mark
- ∠ACB = ∠DCE (given); this is the angle included between the two pairs of equal sides.½ mark
- So ΔACB ≅ ΔDCE (SAS congruence rule).1 mark
- Hence AB = DE (CPCT). Chords that subtend equal angles at the centre are equal.½ mark
Answer to write in the exam
Given: ∠ACB = ∠DCE, C the centre. To prove: AB = ED.
In ΔACB and ΔDCE:
AC = DC (radii)
∠ACB = ∠DCE (given)
BC = EC (radii)
∴ ΔACB ≅ ΔDCE (SAS)
∴ AB = DE (CPCT)
Common mistakes that cost marks
- Using SSS: the chords are what we want to prove equal, so they cannot be used as a given side.
- Writing ASA or AAS: only one angle is known, and it lies between the known sides, so the rule is SAS.
- Forgetting to say “radii of the same circle” as the reason for AC = DC and BC = EC.
How this can come in the exam
In a circle with centre O, ∠AOB = ∠COD = 50° and AB = 4.5 cm. Then CD is
- 2.25 cm
- 4.5 cm
- 9 cm
- 50 cm
Show answer
(B) 4.5 cm
Chords that subtend equal angles at the centre are equal, so CD = AB = 4.5 cm.
Chords AB and CD of a circle with centre O each subtend a right angle at O. The radius is 5 cm. Find AB and CD.
Show answer
In right triangle AOB, AB2 = 52 + 52 = 50, so AB = 5√2 ≈ 7.07 cm (1 mark). ∠COD = ∠AOB, so CD = AB = 5√2 cm (equal angles at the centre give equal chords) (1 mark).Try one yourself
A wheel has 8 equally spaced spokes. The rim points where neighbouring spokes meet the rim are joined. Explain why all 8 joining chords are equal.
Show answer
Neighbouring spokes make 360° ÷ 8 = 45° at the centre. All 8 chords subtend equal angles (45°) at the centre, so they are all equal.
More questions like this
- Show that the triangle formed by a chord and the centre of the circle is isosceles.
- 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.