Equal chords of a circle subtend equal angles at the centre of the circle.
Step-by-step solution
To find: Show that ∠ACB = ∠DCE
Idea: To show two angles in different triangles are equal, show the triangles are congruent with these angles in matching positions. Here all three sides match: two pairs are radii and the third pair is the equal chords.
- In ΔCAB and ΔCDE: CA = CD (radii of the same circle) and CB = CE (radii).1 mark
- AB = DE (given).½ mark
- So ΔCAB ≅ ΔCDE (SSS congruence rule).1 mark
- Hence ∠ACB = ∠DCE (corresponding parts of congruent triangles, CPCT). So equal chords subtend equal angles at the centre.½ mark
Answer to write in the exam
Given: AB = DE, chords of a circle with centre C. To prove: ∠ACB = ∠DCE.
In ΔCAB and ΔCDE:
CA = CD (radii)
CB = CE (radii)
AB = DE (given)
∴ ΔCAB ≅ ΔCDE (SSS)
∴ ∠ACB = ∠DCE (CPCT)
Common mistakes that cost marks
- Writing SAS instead of SSS. No angle is known at the start; all three sides are known to be equal.
- Matching the vertices in the wrong order (e.g. ΔCAB ≅ ΔDCE). The centre must match the centre so that ∠ACB corresponds to ∠DCE.
- Using the theorem for chords of two different circles. The radii are equal only when both chords are in the same circle (or in equal circles).
How this can come in the exam
AB and PQ are equal chords of a circle with centre O. If ∠AOB = 64°, then ∠POQ is
- 32°
- 64°
- 116°
- 128°
Show answer
(B) 64°
Equal chords subtend equal angles at the centre, so ∠POQ = ∠AOB = 64°.
In a circle with centre O, chords PQ and RS are equal. If ∠POQ = (4x − 15)° and ∠ROS = (2x + 27)°, find x and ∠POQ.
Show answer
Equal chords subtend equal angles at the centre, so 4x − 15 = 2x + 27 (1 mark). Hence 2x = 42, x = 21 and ∠POQ = 4 × 21 − 15 = 69° (1 mark).Try one yourself
In a circle with centre O, chords AB, BC and CD are all equal and ∠AOB = 50°. Find ∠AOD (the angle through B and C).
Show answer
∠AOB = ∠BOC = ∠COD = 50° (equal chords). So ∠AOD = 3 × 50° = 150°.
More questions like this
- Chords of a circle that subtend equal angles at the centre are equal.
- 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.)