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.)
Step-by-step solution
To find: Show that AM = BM
Idea: The known result says: the line from the centre to the midpoint of a chord is ⟂ the chord. Its converse says: perpendicular from the centre ⇒ it meets the chord at its midpoint. Two right triangles with equal hypotenuses and a common side are congruent (RHS).
- Join CA and CB. In ΔCMA and ΔCMB: ∠CMA = ∠CMB = 90° (given).½ mark
- Hypotenuse CA = hypotenuse CB (radii of the circle).½ mark
- CM = CM (common side).½ mark
- So ΔCMA ≅ ΔCMB (RHS congruence rule).1 mark
- Hence AM = BM (CPCT): the perpendicular from the centre bisects the chord.½ mark
Check: Another way: by the Baudhāyana–Pythagoras theorem, AM2 = CA2 − CM2 and BM2 = CB2 − CM2; since CA = CB, AM = BM.
Answer to write in the exam
Given: CM ⟂ AB, C the centre. To prove: AM = BM.
In ΔCMA and ΔCMB:
∠CMA = ∠CMB = 90° (given)
CA = CB (radii)
CM = CM (common)
∴ ΔCMA ≅ ΔCMB (RHS)
∴ AM = BM (CPCT), i.e. the perpendicular from the centre bisects the chord.
Common mistakes that cost marks
- Using SAS with CM common and the right angles, plus “AM = BM”: that is the result to be proved, not a given.
- Calling CM a radius. CM goes from the centre to the chord, not to the circle.
- Mixing up the theorem and its converse: here the right angle is given and the midpoint is proved.
How this can come in the exam
OM is the perpendicular from the centre O to a chord AB, and AM = 3.5 cm. The length of AB is
- 3.5 cm
- 7 cm
- 1.75 cm
- 10.5 cm
Show answer
(B) 7 cm
The perpendicular from the centre bisects the chord, so MB = AM = 3.5 cm and AB = 7 cm.
Try one yourself
The perpendicular from the centre of a circle of radius 85 mm to a chord is 13 mm long. Find the length of the chord.
Show answer
Half-chord = √(852 − 132) = √(7225 − 169) = √7056 = 84 mm (the perpendicular bisects the chord). Chord = 168 mm.
More questions like this
- 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.
- Use the Baudhāyana–Pythagoras theorem to show why chords of a circle having the same length are all at the same distance from the centre of the circle.