The line joining the centre of a circle and the midpoint of a chord of the circle is perpendicular to the chord.
Step-by-step solution
To find: Show that CM ⟂ AB
Idea: Prove that the two angles at M are equal; since they sit on a straight line they add up to 180°, so each must be 90°. The equal angles come from congruent triangles CMA and CMB.
- ΔCAB is isosceles with CA = CB (radii), so its base angles are equal: ∠CAB = ∠CBA, that is, ∠CAM = ∠CBM.½ mark
- M is the midpoint of AB, so AM = BM.½ mark
- In ΔCMA and ΔCMB: CA = CB, ∠CAM = ∠CBM, AM = BM. So ΔCMA ≅ ΔCMB (SAS). Hence ∠CMA = ∠CMB (CPCT).1 mark
- But ∠CMA + ∠CMB = 180° (angles on a straight line). So 2∠CMA = 180°, and ∠CMA = ∠CMB = 90°. Thus CM ⟂ AB.1 mark
- (SSS works too: CA = CB, AM = BM and CM is common.)
Answer to write in the exam
Given: C is the centre, M the midpoint of chord AB. To prove: CM ⟂ AB.
CA = CB (radii) ⇒ ∠CAM = ∠CBM (angles opposite equal sides)
AM = BM (M is the midpoint)
In ΔCMA and ΔCMB: CA = CB, ∠CAM = ∠CBM, AM = BM
∴ ΔCMA ≅ ΔCMB (SAS) ⇒ ∠CMA = ∠CMB (CPCT)
∠CMA + ∠CMB = 180° (linear pair) ⇒ ∠CMA = ∠CMB = 90°
∴ CM ⟂ AB
Common mistakes that cost marks
- Assuming ∠CMA = 90° at the start; that is what has to be proved.
- Stopping at ∠CMA = ∠CMB. The key last step is that they form a linear pair, so each is 90°.
- Using RHS congruence, which needs a right angle that is not yet known.
How this can come in the exam
M is the midpoint of a chord AB (not a diameter) of a circle with centre O. Then ∠OMB is
- 45°
- 60°
- 90°
- 180°
Show answer
(C) 90°
The line from the centre to the midpoint of a chord is perpendicular to the chord, so ∠OMB = 90°.
The radius of a circle is 17 cm and the distance from the centre O to the midpoint M of a chord AB is 8 cm. Find AB.
Show answer
OM ⟂ AB, so in right ΔOMA: AM = √(172 − 82) = √(289 − 64) = √225 = 15 cm (1 mark). AB = 2 × AM = 30 cm (1 mark).Try one yourself
In a circle with centre O, M is the midpoint of chord PQ and ∠OPM = 28°. Find ∠POM.
Show answer
∠OMP = 90° (line from centre to midpoint of a chord), so ∠POM = 180° − 90° − 28° = 62°.
More questions like this
- 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? - Chords of a circle having the same length are all at the same distance from the centre of the circle.