The centre-to-midpoint line is perpendicular, and the converse (Theorems 4–5)

What to assume they know

• Chords of equal length cut off equal central angles, and the converse (Theorems 2–3) — the isosceles triangle a chord makes with the centre, and the SSS and SAS arguments built on it
• Total rotational symmetry, and why every diameter is an axis of reflection — reflection in a diameter, and the fold that produces a crease
• That angles making a straight line add to 180°
• SAS congruence, and reading off corresponding angles
• Perpendicular bisector of a segment, and altitude of a triangle
• The Pythagorean relation in a right triangle — the chapter calls it the Baudhāyana–Pythagoras theorem

What they should be able to do

• Explain why the segment from the centre to a chord's midpoint is the axis of symmetry of the triangle the chord makes with the centre
• Prove that joining the centre to a chord's midpoint gives a segment perpendicular to the chord
• Identify the step where the congruence turns two equal angles into two right angles, and say why a straight line is needed for it
• State the converse and prove it, and say why it needs an argument of its own
• Explain why any chord's perpendicular bisector is forced to run through the centre
• Use the two results together to locate a chord's midpoint or its distance from the centre
• Show that in a triangle inscribed in a circle with two equal sides, the altitude to the third side runs through the centre
• Compute the separation of two parallel chords of stated lengths in a circle of stated radius, on the same side and on opposite sides

Where it usually goes wrong

• "Perpendicular and bisecting are the same, so one theorem covers both." They are equivalent for a line through the centre, and that is a result, not a definition. A chord's perpendicular that does not pass through the centre does not bisect it — indeed there is no such perpendicular through the centre other than the one, which is what makes the pair of theorems worth stating.
• "The right angle in Fig. 5.12 is given." In Theorem 4 it is the conclusion. The chapter's figure deliberately omits the tick. When the same figure is reused for the converse the right angle becomes the given.
• "Two equal angles must be 90°." Only if they also make a straight line. This is the step where the proof actually happens and it is the step students skip.
• "Any segment drawn from the centre down to a chord halves it." Only the perpendicular one does. Draw a slanted segment from the centre to a chord and the two pieces are unequal.
• "The perpendicular bisector of a chord is a diameter." The line containing it cuts the circle in two points, and the chord joining those is a diameter. The bisector itself is a line, not that chord; the chapter keeps diameter for the chord.
• "Same side or opposite side does not matter for parallel chords." It decides whether the two distances add or subtract. Exercise Set 5.3 Q3 says opposite and End-of-Chapter Q18 says same side, and they are the same problem with different answers. This is the single most common arithmetic loss on this topic.
• "The fold argument replaces the proof." The chapter is explicit at the top of p. 103 that demonstrations on examples do not establish a general claim. Use the fold to make the reflection visible; keep the congruence.

Questions to check understanding

• Prove that the segment joining the centre to a chord's midpoint is perpendicular to the chord
• Prove the converse — that a perpendicular dropped from the centre halves the chord
• Prove that a chord's perpendicular bisector must run through the centre (the form End-of-Chapter Q5, p. 114 takes)
• Two parallel chords of given lengths in a circle of given radius: find their separation, on the same side and on opposite sides
• Given a radius and a chord's distance from the centre, find the chord length, and the reverse
• Show that the altitude from the apex of an inscribed isosceles triangle passes through the centre
• Construction: given a drawn circle with no marked centre, locate the centre and justify each step
• One-mark: a chord is 8 cm long in a circle of radius 5 cm — how far is it from the centre?

Examples worth working on the board

The chapter prints no answers, so every value marked verified is worked out here on the chapter's own stated inputs.

• Fig. 5.12 (p. 101). Circle with centre C. Chord AB drawn with A on the upper left and B on the right; M marked on the chord as its midpoint; the segments CA, CB and CM all drawn, so the chord and the centre make a triangle that CM splits in two. The figure carries no right-angle tick — the right angle is the conclusion, not a given, and drawing it in would beg the question. Note also that the same figure is reused for the converse, where the right angle is given: the hint at Exercise Set 5.3 Q1 says so explicitly.
• The Theorem 4 argument, as data. CA = CB = r, so the base angles at A and B are equal. AM = BM, because M is the midpoint. That is two sides and the included angle in each of triangles CMA and CMB, so SAS gives the congruence, and the two angles at M come out equal. Those two angles together make the straight line AB, so they sum to 180°, so each is 90°.
• The converse, as the chapter stages it (Exercise Set 5.3 Q1, p. 101). Given the two angles at M are each 90°, show AM = BM. The printed hint names Fig. 5.12 and states the given. Verified: CA = CB = r and CM is shared, with right angles at M, so RHS congruence gives triangles CMA and CMB congruent and hence AM = BM. The chapter then states the result as Theorem 5 on p. 102 and credits it to that question — so the exercise is load-bearing, not optional.
• Exercise Set 5.3, Q2 (p. 101). Inputs: a triangle ABC with AB = AC and all three vertices on a circle; show that the altitude dropped from A onto BC runs through the centre. Verified: AB = AC puts A on the perpendicular bisector of BC; the centre — call it O, since the printed question leaves it unnamed — is on that bisector too, because OB and OC are both radii; and in a triangle with AB = AC the altitude from A to BC is that same perpendicular bisector. So all three coincide.
• Exercise Set 5.3, Q3 (p. 101). Inputs: two parallel chords, 6 cm and 8 cm, on opposite sides of the centre; radius 5 cm; find the distance between the chords' midpoints. Verified: half-chords 3 and 4, so distances from the centre √(25 − 9) = 4 and √(25 − 16) = 3, and because the chords are on opposite sides the midpoints are 4 + 3 = 7 cm apart. On the same side the separation would be 4 − 3 = 1 cm, and End-of-Chapter Q18 on p. 115 uses the same-side version with chords of 10 cm and 24 cm.
• End-of-Chapter Q5 (p. 114). Prove that a chord's perpendicular bisector has to run through the centre. Verified: the centre is equidistant from the chord's two ends, both being radii, so it lies on that perpendicular bisector. This is the form the results take when a construction needs them, and it is the reason fold-twice finds a paper circle's centre.
• Numbers for section 10's build-up (not in the book, not printed): radius 5. Verified: a chord 4 cm from the centre is 6 cm long; a chord 3 cm from the centre is 8 cm long; a chord through the centre is 10 cm long and its distance is 0. Use these to show that fixing the distance fixes the length, before §5.6 says so as a theorem.

Figures to have open

• Fig. 5.12 redrawn twice: once without a right-angle tick, for Theorem 4, and once with the tick given, for the converse. The chapter reuses one figure for both roles (p. 101 and its Exercise Set 5.3 hint) and the reuse is exactly where students lose the thread.
• A fold overlay: the paper-disc crease of Fig. 5.13, p. 102, laid against the drawn centre-to-midpoint segment. The chapter's own photographs are on p. 102 and belong to Length and distance from the centre are the same fact twice (Theorems 6–8); borrowing one panel here is worth it.
• A radius-5 circle carrying both chords of Exercise Set 5.3 Q3 on opposite sides of the centre, with the two distances and the total marked, plus a same-side variant beside it. Standard schematic; the chapter draws no figure for this question.
• A counter-figure for the misconceptions: a slanted segment from the centre to a chord, with the two unequal pieces measured. Standard schematic; not in the book.
• No photograph is strictly required.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics, printed Chapter 5, "I’m Up and Down, and Round and Round", §5.5, printed heading "Midpoints and Perpendicular Bisectors of Chords", from the foot of p. 100 through p. 101 and onto the top of p. 102 where Theorem 5 is stated.
• Fig. 5.12 with its caption, p. 101 — the section's only figure.
• Exercise Set 5.3, Q1–Q3, p. 101. Q1 carries a printed hint and is the source of Theorem 5.
• Theorem 5 as stated, p. 102, with the sentence crediting it to Exercise Set 5.3 Q1.
• Related items elsewhere in the chapter: End-of-Chapter Q5 and Q18, pp. 114–115; the Baudhāyana–Pythagoras theorem is named at Exercise Set 5.4, p. 104.
• Chapter Summary, p. 117 — the bullet pairing the centre-to-midpoint result with its converse.