Cyclic quadrilaterals: opposite angles sum to 180°, and the converse (Theorems 11–12)

What to assume they know

• Equal angles on the same side force concyclicity (Theorem 10) — Theorem 10, concyclicity by elimination, and the words cyclic quadrilateral and 4-gon
• 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, including readings past 180°
• Equal angles in the same segment: the arc looks the same from every point beyond it — equal angles from points off the same arc
• Three points not in a line: exactly one circle (Theorem 1) — Theorem 1, exactly one circle through three points off a line
• That a complete turn about a point is 360°, and what a reflex angle is
• The exterior-angle theorem, and proof by contradiction

What they should be able to do

• Define cyclic quadrilateral and use the chapter's alternative word 4-gon
• Identify, for a given corner of a cyclic quadrilateral, which arc it stands off
• Explain why one of the two central angles involved has to be read as a reflex angle
• Derive the 180° total from the two arcs making up a complete turn
• Find the remaining angles of a cyclic quadrilateral given one or two of them
• Solve for an unknown when two opposite angles are given algebraically
• Decide whether a stated set of four angles can belong to a cyclic quadrilateral, and say which theorem settles it
• Prove the converse by eliminating the case where the fourth vertex is off the circle
• Derive the exterior-angle property of a cyclic quadrilateral from the opposite-angle result

Where it usually goes wrong

• "Any quadrilateral has opposite angles adding to 180°." Only a cyclic one. A general quadrilateral's four angles total 360°, which says nothing about the pairs. This is the error the p. 113 exercise is built to expose.
• "Adjacent angles add to 180°." It is the opposite pair. Students transfer the parallelogram fact.
• "A central angle cannot be reflex, so the proof must be wrong." The reflex reading is essential; without it the two arcs would not account for the whole circle. The chapter marks it with a dotted arc in Fig. 5.28 and names it in the text, and this is where §5.7's swept-angle definition finally pays for itself.
• "Checking that the four angles total 360° checks that the quadrilateral is cyclic." Every quadrilateral totals 360°. Both opposite pairs have to be checked.
• "The converse is obvious once you have the theorem." It is not, and the chapter proves it by contradiction over two cases — leaving one of them to the reader. A statement and its converse are different claims; this chapter makes that point four separate times.
• "Two supplementary opposite angles in a quadrilateral means both pairs are supplementary." In fact it does follow, since the four total 360°.
• "A cyclic parallelogram is possible." Only if it is a rectangle: opposite angles of a parallelogram are equal, and if they are also supplementary each is 90°. That is starred Q14 on p. 115.
• "The exterior angle equals the adjacent interior angle." It equals the interior angle at the opposite corner. The adjacent one is its supplement.

Questions to check understanding

• Given one or two angles of a cyclic quadrilateral, find the rest (the form End-of-Chapter Q7, p. 114 takes)
• Given two opposite angles algebraically, solve for the unknown and state both angles (End-of-Chapter Q8, p. 114)
• Decide whether four stated angles can belong to a cyclic quadrilateral, with a reason
• Prove that opposite angles of a cyclic quadrilateral are supplementary
• Prove the converse
• Prove that an exterior angle of a cyclic quadrilateral equals the interior angle at the opposite corner
• Show that a parallelogram inscribed in a circle must be a rectangle
• Find the area of a cyclic quadrilateral with two pairs of adjacent equal sides
• Justify the 180° result from the four-isosceles-triangle figure

Examples worth working on the board

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

• Fig. 5.28 (p. 112). Theorem 11's figure. Circle with centre O marked inside. Four points on the circle make the 4-gon: A on the left, B at the top, C at the lower right, D at the lower left, with the four sides drawn. The radii OB and OD are drawn, and O carries two angle marks, both of which the printed argument uses: a dotted arc sweeping the reflex angle BOD, which is the one the chapter names in words, and a double solid arc on the non-reflex angle BOD. The dotted mark serves the angle at A, the double-arc mark the angle at C, so a redraw that keeps only the dotted one throws away half the figure's apparatus. Angle marks sit at A and at C, the two corners the theorem is about. Given: A, B, C, D are the corners of a cyclic 4-gon with centre O. To show: the angles BAD and BCD total 180°.
• The argument, as data. Consider the arc from B to D that passes through C. The corner A is on the circle and off that arc, so the angle BAD is half the central angle of that arc — and going from OB to OD the way that passes C sweeps past a straight angle, so that central angle is the reflex one. Now consider the arc from B to D that passes through A: the corner C is off it, so the angle BCD is half that arc's central angle, the non-reflex one. The two central angles between them are one complete turn, 360°. Half of 360° is 180°. The step to slow down on is the choice of arcs — each corner is paired with the arc it does not sit on.
• The other pair, for completeness. Verified: the same argument with the arcs on A and C instead of B and D gives that the angles ABC and ADC also total 180°. The chapter states the conclusion for opposite angles generally after working one pair.
• The Exercise (§5.8, p. 113). Inputs: a cyclic quadrilateral with angle A = 80°, angle B = 110°, angle C = 100°, angle D = 70°; can it be drawn? Verified: yes. A + C = 180° and B + D = 180°, and the four total 360° as any quadrilateral must; so by Theorem 12 those four angles do belong to a cyclic quadrilateral.
• Fig. 5.29 (p. 113). Theorem 12's figure, for the case the chapter works. A circle is drawn through A, D and B — A at the upper left, B at the upper right, D on the left — with the centre O marked. C is placed below and outside the circle, and the side CD meets the circle at E, which is lettered just above C — E lies between D and C on that segment, so nothing is produced or extended. With D on the circle and C outside it, DC dips through the interior and leaves once, and E is where it leaves; the page words it the same way. The segment from B to E is drawn dashed. Given: ABCD is a 4-gon with both pairs of opposite angles totalling 180°. To show: ABCD is cyclic.
• The converse argument, as data. Suppose ABCD is not cyclic. A, D and B are not in a line, so Theorem 1 gives exactly one circle through them; if C is off it, C is either outside or inside. Take the outside case, which is the one the chapter works. Let E be where CD meets that circle. Then ABED is a cyclic 4-gon, so Theorem 11 gives that the angles BAD and BED total 180°. But the given says the angles BAD and DCB total 180° too. So the angles BCD and BED are equal — which is impossible, because BED is exterior to triangle BEC and so strictly exceeds the angle at C. Contradiction; C must be on the circle.
• The case the chapter leaves out (p. 113). The chapter says explicitly that it handles only the first case and leaves the other to the reader. Verified: with C inside the circle, the line DC extended meets the circle at a point E beyond C, and now the angle BCD is the exterior angle of the small triangle and so exceeds the angle BED; the same chain gives the same impossibility. Flag that the explanation is supplying this — it is not printed.
• End-of-Chapter Q7 (p. 114). Inputs: cyclic ABCD with angle A = 75° and angle B = 110°. Verified: angle C = 105° and angle D = 70°.
• End-of-Chapter Q8 (p. 114). Inputs: PQRS inscribed in a circle, angle P = (2x + 10)° and angle R = (3x − 20)°. Verified: P and R are opposite, so 5x − 10 = 180, giving x = 38, angle P = 86° and angle R = 94°.
• Exercise Set 5.6 Q3 and Fig. 5.26 (p. 111). The chapter's only in-text numerical cyclic-quadrilateral exercise, and no other brief carries it. Fig. 5.26 is a plain cyclic 4-gon lettered A, D, C, B, with 100° printed at D and an italic x at B — the unknown, which is present on the page. Verified: D and B are the opposite pair, so Theorem 11 gives x = 80°. Worth using as the first, easiest application before the end-of-chapter items.
• End-of-Chapter Q10 (p. 114). Inputs: a cyclic quadrilateral with sides 5, 5, 12 and 12 units; find its area. Verified: the question gives no cyclic order, so the kite is the reading being taken rather than something forced — 5, 12, 5, 12 would be a 5 by 12 rectangle and fits the wording just as well. Nothing downstream changes: Brahmagupta's formula is symmetric in the four sides, so s = 17 gives an area of √(12·12·5·5) = 60 either way, with a 13-unit diagonal that is a diameter in both. Take it as a kite and say that is the reading; being cyclic, its two remaining opposite angles are equal and supplementary, hence each 90°; so it is two right triangles with legs 5 and 12, area 60 square units, and its long diagonal is 13 — a diameter. This is the hardest unstarred item in the set.
• End-of-Chapter Q21 (p. 115). Inputs: a cyclic quadrilateral ABCD, and the claim that an exterior angle at any corner equals the interior angle at the opposite corner. Verified: extend a side through a corner; the exterior angle and the interior angle there make a straight line, so they total 180°, and the interior angle and its opposite also total 180° — so the exterior angle equals the opposite interior angle. See the note below about the printed lettering of this question.
• Fig. 5.31 and End-of-Chapter Q26 (p. 116). A second route to the theorem in the configuration Fig. 5.31 draws, with O inside the 4-gon; scope it that way, because splitting each corner angle into two base angles depends on it. If the circumcentre falls outside, a corner angle becomes a difference rather than a sum and the count has to be redone — which is exactly what the starred-Q11 bullet below is about. Supplied as a figure. Circle with centre O; the 4-gon ABCD shaded, with A at the upper left, B at the lower left, C at the lower right and D on the right. All four radii OA, OB, OC and OD are drawn and carry tick marks showing them equal, and four angles are lettered inside the figure: p at A, q at B, u at C and v at D. Verified, and read the four letters carefully: each takes one base angle from a different triangle, in rotation, not two from each of two triangles. On the printed page, p spans AB to AO, q spans BO to BC, u spans CO to CD, v spans DO to DA. Since each of the four triangles is isosceles on two radii, its two base angles match, and the corner angles come out as sums of two different letters: A = p + v, B = p + q, C = q + u, D = u + v. The eight base angles total 360°, because the four triangles total 720° and the four angles at O total 360°, so p + q + u + v is 180° — and then each pair of opposite corners sums to 180° immediately, with no equality argument needed. (Reading p and q as two base angles of the same triangle would force them equal, which the four distinct letters rule out.) The question asks the student to construct this; the chapter prints no working.
• End-of-Chapter starred Q11 (pp. 114–115). Inputs: a cyclic quadrilateral; decide without drawing the circumcircle whether the circumcentre falls inside or outside it. Suggested criterion, mine and not the chapter's: the centre falls inside exactly when every side of the 4-gon subtends an acute angle at the opposite corner, because a side subtending a right or obtuse angle cuts off an arc of at least a semicircle and pushes the centre to the far side of that side. Offer it as the explanation's proposal and say so.

Figures to have open

• Fig. 5.28 redrawn with both marks at O kept — the dotted reflex arc and the double solid arc on the non-reflex angle — and the two arcs traced separately. This is the chapter's own figure (p. 112) and the reflex marking is the single most important mark on it; extraction shows nothing of it.
• Fig. 5.29 redrawn with C clearly outside the circle, E on the circle between D and C, and the dashed segment BE kept dashed. Chapter's own figure (p. 113). A version with C inside, for the case the chapter omits, has to be drawn fresh.
• Fig. 5.31 redrawn keeping every tick mark and all four angle letters. Chapter's own figure (p. 116); the ticks are the argument and a redraw that drops them makes Q26 unanswerable.
• A full-turn movement for section 6: the two central angles laid end to end filling 360°. Standard schematic; not in the book.
• A pair-checking panel for section 8: the four printed values against a non-cyclic quadruple that also totals 360°. Standard schematic; the chapter supplies only the cyclic set.
• 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", §5.8, printed heading "Concyclicity of Points". The cyclic quadrilateral is named in the middle of p. 112; Theorem 11 with its figure and the start of its argument are on p. 112; the argument finishes on p. 113, followed by the Exercise, Theorem 12 and its argument.
• Fig. 5.28, p. 112. Fig. 5.29, p. 113. Fig. 5.31, p. 116.
• The Exercise on the four stated angles, p. 113. The sentence saying the converse also holds, p. 113. The sentence leaving the second case of Theorem 12 to the reader, p. 113.
• End-of-Chapter Exercises Q7, Q8, Q10 and starred Q11, p. 114; Q21 and starred Q14, p. 115; starred Q26, p. 116.
• The chapter's closing paragraph on p. 114 names the properties covered and points forward to next year's work.
• Chapter Summary, p. 117 — the final bullet, pairing the opposite-angle result with its converse.