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.
Answer: For equal chords AB and CD of a circle with centre O: OA = OC, OB = OD (radii) and AB = CD (equal bases), so ΔOAB ≅ ΔOCD by SSS.
Step-by-step solution
Given: A circle with centre O; Isosceles triangles OAB and OCD formed by chords AB and CD; AB = CD
To find: Show that ΔOAB ≅ ΔOCD
To find: Show that ΔOAB ≅ ΔOCD
Idea: “Such isosceles triangles” are triangles formed by a chord and the centre of the circle: their two equal sides are radii. With equal bases, all three pairs of sides match, so SSS applies.
- Let AB and CD be chords of a circle with centre O and AB = CD. The triangles are ΔOAB and ΔOCD. In them: OA = OC and OB = OD (all radii of the same circle).1 mark
- AB = CD (given equal bases). So ΔOAB ≅ ΔOCD (SSS congruence rule).1 mark
- As a result, the corresponding angles are also equal; in particular ∠AOB = ∠COD (equal chords subtend equal angles at the centre).
The two isosceles triangles have OA = OC and OB = OD (radii) and equal bases AB = CD, so they are congruent by SSS.
Answer to write in the exam
Given: chords AB = CD of a circle with centre O. To prove: ΔOAB ≅ ΔOCD.
OA = OC (radii)
OB = OD (radii)
AB = CD (given)
∴ ΔOAB ≅ ΔOCD (SSS)
Common mistakes that cost marks
- Applying the result to chords of two different circles. The equal sides are radii, so the circles must be the same (or have equal radii).
- Writing “SAS” without naming an included angle; no angle is given here.
- Listing the sides in a non-matching order, e.g. OA = OD and OB = OC, and then writing ΔOAB ≅ ΔOCD. The order of vertices must follow the matching.
How this can come in the exam
MCQ (1 mark)
AB and CD are equal chords of a circle with centre O, and ∠OAB = 40°. Then ∠OCD is
- 20°
- 40°
- 50°
- 100°
Show answer
(B) 40°
ΔOAB ≅ ΔOCD (SSS), so ∠OCD = ∠OAB = 40°.
Try one yourself
A chord of length 6 cm is drawn in a circle of radius 5 cm, and another chord of length 6 cm in a circle of radius 4 cm. Are the two triangles formed with the centres congruent? Why?
Show answer
No. The bases are equal (6 cm), but the equal sides are radii of different lengths (5 cm and 4 cm), so SSS does not hold.
More questions like this
- 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.
- Take a paper circle. Fold the circle from the boundary, inwards. Open the fold. The crease is now a chord (see the figure B). Now fold the paper again, so that the end points of the chord meet. Open the fold (see the figure C).
Measure the lengths of the parts into which the chord is divided. The chord gets bisected where the folds intersect. Measure the angle between the creases. The crease of the second fold is along the perpendicular from the centre to the chord. Measure the distance from the centre to the midpoint of the chord. It is the distance from the centre to the chord.
Now draw another chord of the same length. How will you do this? We will let you figure this out yourself. Join the centre to the midpoint of the new chord and measure its length. Is it the same as distance from the centre to the first chord?