The Corollary: a diameter stands on a right angle wherever you take the point

What to assume they know

• Major and minor arcs, and why an arc's central angle is double what it subtends on the circle (Theorem 9) — the doubling relation and the swept-angle definition of an arc's central angle
• Equal angles in the same segment: the arc looks the same from every point beyond it — that the angle is the same at every point of the circle off the arc
• Three points not in a line: exactly one circle (Theorem 1) — the circumcentre, and that for a right-angled triangle it sits at the hypotenuse's midpoint
• Diameter, and that the two radii of a diameter point in opposite directions
• A straight angle is 180°
• That a triangle's angles add to 180°

What they should be able to do

• Say what a corollary is, in the sense the chapter defines
• State the right-angle result for a diameter, and identify which arc has to be chosen for the argument to work
• Explain why the swept angle for a semicircular arc is 180° and not 0°
• Derive the right angle in one step from the doubling relation
• Explain why the answer does not change as the third point moves round the circle
• Read the result backwards: given a right angle standing on a segment, place the point on a circle with that segment as diameter
• Connect the result to the circumcentre of a right-angled triangle sitting at the hypotenuse's midpoint
• Reconstruct the same result by the two-isosceles-triangle route the end-of-chapter figure sets up

Where it usually goes wrong

• "The right angle is at the centre." It is at the point on the circle. The angle at the centre is the straight angle, 180°, which is what gets halved.
• "The third point has to be at the top of the semicircle." It can be anywhere on the circle other than the diameter's own ends. Fig. 5.24 draws it at the top and students read that as a requirement.
• "It only works for a semicircle drawn as a half-disc." The relevant object is the arc on one side of the diameter. Nothing has to be cut or shaded.
• "A corollary is a small theorem." The chapter's definition is about derivation, not importance: a corollary follows immediately from something already proved. This one is among the most used results in the whole chapter.
• "Choosing either arc gives the same argument." Choose the arc containing the third point and the configuration no longer matches the doubling relation's requirement. The chapter says explicitly which arc it is taking; an explanation that glosses over that step leaves the argument unjustified.
• "Any question mentioning a diameter is answered by 90°." End-of-Chapter Q20 is built to punish this. Check which chord the asked-about angles actually stand on.
• "The converse is obvious." Going backwards — a right angle placing the point on a specific circle — needs the concyclicity machinery of §5.8. It is true and it is not free.

Questions to check understanding

• Given a diameter and a point on the circle, state the angle and justify it (the form End-of-Chapter Q6, p. 114 takes)
• Justify the right angle in a semicircle from a supplied ticked figure (the form End-of-Chapter Q24, p. 116 takes)
• Given a quadrilateral inscribed in a circle with one side a diameter, identify which angles are right angles and which are merely equal
• Given a right angle standing on a segment, locate the circle the vertex lies on
• Find the circumradius of a right-angled triangle from its legs
• Show that a rectangle inscribed in a circle has its diagonals meeting at the centre
• Show that a parallelogram inscribed in a circle must be a rectangle
• Define corollary and give the chapter's example

Examples worth working on the board

The chapter prints no answers, so anything marked verified is worked out here, not the book's.

• The Corollary's box (§5.7.1, p. 110). Two things sit here and both matter. The corollary itself is printed in italics above, and below it a separate ruled box defines what a corollary is. The definition box is unusual — the chapter is teaching a piece of mathematical vocabulary about the structure of results, not about circles.
• Fig. 5.24 (p. 110). Circle with centre C. B on the left and A on the right, with the segment from B through C to A drawn — so BA is a diameter and C is its midpoint as well as the circle's centre. D sits at the top of the circle, and the chords DB and DA are drawn, making a triangle on the diameter. The angle to be found is the one at D.
• The one-step argument, as data. Choose the arc from A to B that does not contain D. Sweeping a radius from CA round that arc to CB turns it through a straight angle, so the arc's central angle is 180°. The doubling relation says D's angle is half of that. Verified: 90°. Note: the whole argument is one substitution, and the only step a student can get wrong is choosing the other arc, which would give 180° at the centre going the wrong way round and no sensible reading at D.
• Why the answer is immobile. Verified: the arc, and therefore its central angle, does not change when D moves; so neither does D's angle. This is Equal angles in the same segment: the arc looks the same from every point beyond it's invariance applied to one particular arc, and it is worth naming as such rather than re-deriving.
• The backwards reading, worked. Take a segment AB of length 10 and a point P with the angle APB equal to 90°. Verified: P lies on the circle of radius 5 whose centre is AB's midpoint, so PA² + PB² = 100 for every such P; e.g. PA = 6 and PB = 8. This gives the explanation a concrete family — added numbers, not the chapter's.
• Fig. 5.7 (p. 97). The right-angled triangle whose circumcentre sits at the hypotenuse's midpoint, with all three perpendicular bisectors drawn and labelled. This figure and the Corollary are the same fact read from opposite ends: if the centre is the hypotenuse's midpoint then the hypotenuse is a diameter, and the right angle at the third vertex is the Corollary. The chapter prints them thirteen pages apart and never says they are connected. Connecting them is one of the more valuable things the explanation can do.
• Fig. 5.30 and End-of-Chapter Q24 (p. 116). The figure gives the alternative route. A semicircular arc stands on a horizontal diameter; the two ends of the diameter carry dots but no letters — only the apex A and the centre O are lettered. The triangle on the diameter is shaded, the segment OA is drawn dashed, and tick marks are set on the left half of the diameter, on the right half and on OA, marking all three as equal. Two angles are lettered inside the triangle: a at the left base vertex and b at the right. Verified: the ticks make two isosceles triangles, so the angle at A splits into a piece equal to a and a piece equal to b; the triangle's three angles then give a + (a + b) + b = 180°, hence a + b = 90°, which is the angle at A. The question asks the student to build this argument; the chapter prints no working.
• End-of-Chapter Q6 (p. 114). Inputs: AB is a diameter and C is on the circumference; give the angle ACB with reasoning. Verified: 90°. This is the Corollary asked directly and it is the commonest one-mark form.
• End-of-Chapter Q20 (p. 115). Inputs: a quadrilateral MNOP inscribed in a circle with MN a diameter; what can be said about the angles MOP and MNP? Verified: they are equal — both stand on the chord MP, and O and N are on the same side of it, so Equal angles in the same segment: the arc looks the same from every point beyond it applies. Worth flagging as a trap: the question mentions a diameter, so students reach for 90° and answer the wrong question. The Corollary does apply here, but to the angles MON and MPN, not to the pair asked about.
• End-of-Chapter starred Q14 and Q15 (p. 115). Q14: show that among parallelograms only a rectangle will fit inside a circle with its corners on it. Q15: show that an inscribed rectangle's diagonals cross at the centre. Verified: Q15 is the one that runs through this Corollary, in its converse direction — a rectangle's diagonal subtends a right angle at each of the other two vertices, so each diagonal is a diameter and the crossing point is the centre. Q14 does not use the Corollary at all, in either direction: a parallelogram's opposite angles are equal, a cyclic 4-gon's are supplementary, so each is 90°, which is Theorem 11 and nothing more. Do not present the pair as a matched use of one result.

Figures to have open

• Fig. 5.24 redrawn with the third point shown step by step round the circle and the right angle persisting. This is the chapter's own figure (p. 110) and the motion is what carries section 6; the printed figure is static and shows one position.
• A two-arc figure for section 3: the diameter with each of its two arcs traced separately, and the chosen one marked. Standard schematic; the chapter states the choice in words and does not draw it.
• Fig. 5.7 and Fig. 5.24 placed side by side. These are the chapter's own figures from pp. 97 and 110; the pairing is not in the book and is the point of section 8.
• Fig. 5.30 redrawn keeping every tick mark and both angle letters. This is the chapter's own figure (p. 116) and the ticks are the argument — a redraw that drops them leaves the question unanswerable. Note that the diameter's ends are unlettered in the book; the explanation will need to letter them and should say it is doing so.
• An MNOP figure for section 10 with the diameter and both angle pairs marked differently. Standard schematic; the question on p. 115 supplies no figure.
• No photograph is needed.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics, printed Chapter 5, "I’m Up and Down, and Round and Round", §5.7.1. The Corollary, the ruled box defining the word, and the argument all sit on p. 110 together with Fig. 5.24.
• Fig. 5.24, p. 110. Fig. 5.7 with its caption, p. 97. Fig. 5.30, p. 116.
• End-of-Chapter Exercises Q6, p. 114; Q20 and starred Q14 and Q15, p. 115; Q24, p. 116.
• Chapter Summary, p. 117 — the bullet on the angle a diameter makes at a point of the circle.
• The circumcentre material this topic reconnects to is §5.3, p. 97; see Three points not in a line: exactly one circle (Theorem 1).