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NCERT Solutions for Class 9th Maths Chapter 5 Exercise Set 5.3 — Midpoints and Perpendicular Bisectors of Chords

Book page 101 Updated on2026-09-08

Q1.
Can you explain why the converse to Theorem 4 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 Fig. 5.12. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM.)
Answer

Given: circle with centre C, chord AB, and CM ⊥ AB with M on AB. To show: AM = BM.

In ΔCMA and ΔCMB:
∠CMA = ∠CMB = 90° (given — CM is perpendicular to AB)
CA = CB = r (radii — these are the hypotenuses)
CM = CM (common side)
By the RHS congruence rule, ΔCMA ≅ ΔCMB
∴ AM = BM (corresponding sides) — CM bisects the chord AB

Alternative, using Baudhāyana–Pythagoras:

AM² = CA² − CM² and BM² = CB² − CM²
CA = CB = r, and CM is the same in both
⇒ AM² = BM² ⇒ AM = BM (lengths are positive)
Why RHS and not SSS: we are given a right angle and the two hypotenuses, and we do not yet know the third pair of sides — that is precisely the RHS situation. Note the contrast with Theorem 4, where we were given the midpoint and had to produce the right angle; there SAS was the right tool.
Q2.
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.
Answer

Given: ΔABC inscribed in a circle with centre O, and AB = AC. Let AD be the altitude from A to BC. To show: O lies on AD.

Step 1 — the altitude from A is the perpendicular bisector of BC. In ΔABD and ΔACD: AB = AC (given), AD common, ∠ADB = ∠ADC = 90°. By RHS, ΔABD ≅ ΔACD, so BD = DC. Thus AD is perpendicular to BC and passes through its midpoint D — it is the perpendicular bisector of BC.

Step 2 — the perpendicular bisector of a chord passes through the centre. BC is a chord, and OB = OC (radii), so O is equidistant from B and C. Every point equidistant from B and C lies on the perpendicular bisector of BC. Hence O lies on that line.

Perpendicular bisector of BC = line AD (Step 1)
O lies on the perpendicular bisector of BC (Step 2)
∴ O lies on AD — the altitude from A passes through the centre
A B C O D
AD bisects BC at right angles, so it is the perpendicular bisector of the chord BC and must contain the centre O.
Tip: for an isosceles triangle the altitude, the median, the angle bisector from the apex and the perpendicular bisector of the base are all the same line — and here that line is also a diameter of the circumcircle.
Q3.
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.
Answer

Distance = 7 cm.

Drop perpendiculars from the centre O to each chord. By Theorem 5 each perpendicular bisects its chord, so its foot is the midpoint of that chord.

Chord AB = 6 cm ⇒ half-chord = 3 cm
d₁² = r² − 3² = 5² − 3² = 25 − 9 = 16 ⇒ d₁ = 4 cm

Chord PQ = 8 cm ⇒ half-chord = 4 cm
d₂² = r² − 4² = 5² − 4² = 25 − 16 = 9 ⇒ d₂ = 3 cm

The chords are on opposite sides of O, and both perpendiculars lie along the same line through O.
Distance between midpoints = d₁ + d₂ = 4 + 3 = 7 cm
O 6 cm 8 cm 4 3
Opposite sides of the centre: the two distances add. Note the longer chord (8 cm) is the nearer one (3 cm).
Tip: "opposite sides" ⇒ add the distances; "same side" ⇒ subtract them. Getting this the wrong way round is the commonest slip in these questions.
Sanity check: Theorem 8 says the longer chord must be closer to the centre. Here the 8 cm chord is 3 cm away and the 6 cm chord is 4 cm away — exactly as predicted.
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