Equal angles on the same side force concyclicity (Theorem 10)

What to assume they know

• Three points not in a line: exactly one circle (Theorem 1) — Theorem 1, exactly one circle through three points off a line
• Equal angles in the same segment: the arc looks the same from every point beyond it — that two points of a circle on one side of a chord give the same angle
• Off the circle the angle changes: points inside and outside compared (Fig. 5.25) — that off the circle the angle changes, and in which direction
• The exterior-angle theorem, and that an exterior angle strictly exceeds either remote interior angle
• Proof by contradiction: assume the opposite, derive something impossible
• That a quantity cannot be strictly greater than itself

What they should be able to do

• State what concyclic means, and say why four points is the first interesting case
• Explain why the four-point question reduces to a three-plus-one question
• State the two conditions Theorem 10 requires, and give an example showing that dropping the same-side condition breaks it
• Build the circle through three of the points, and say which earlier theorem guarantees it is the only one
• Enumerate the two ways the fourth point could be off that circle, and set up the auxiliary point E in each
• Run the exterior-angle inequality on the small triangle and reach the contradiction
• Explain why proving that something is impossible establishes the positive claim
• Define cyclic quadrilateral and 4-gon as the chapter uses them

Where it usually goes wrong

• "Any four points lie on some circle." Three points off a line do; a fourth is a genuine constraint. Students who absorbed "three points, one circle" often expect the pattern to continue.
• "Draw the circle through all four and you are done." You cannot draw what you are trying to prove exists. The circle you are entitled to is the one through three of them, and that is why the argument has the shape it has.
• "Equal angles anywhere will do." They must be on the same side of the segment. Opposite sides give supplementary angles, and Theorem 10 says "same side" for exactly that reason.
• "A contradiction only shows that particular drawing is wrong." It shows the supposition is impossible, in every drawing. That is what makes elimination a proof rather than a check.
• "Two cases means two theorems." The two cases are the two ways one point can be off a circle. They are exhaustive, which is why eliminating both settles the matter.
• "E is one of the four points." E is constructed only to make a triangle for the exterior-angle step, and it disappears from the conclusion. Students often try to carry it into the answer.
• "Theorem 1 is just background." Theorem 1 supplies uniqueness, and uniqueness is what makes the contradiction possible. Without it, "D is not on this circle" would be unremarkable.

Questions to check understanding

• Prove that if a segment subtends equal angles at two points on the same side of it, the four points are concyclic
• Given a figure with two equal marked angles on the same side of a segment, state what follows and name the theorem used
• Exercise Set 5.6 Q2(iii), p. 111, as written
• Explain why the same-side condition is needed, with a counter-example
• Explain why the proof needs Theorem 1
• Identify the two cases in the proof and say why they are exhaustive
• Define concyclic and cyclic quadrilateral
• Given three points and a stated angle at a fourth, decide whether all four are concyclic

Examples worth working on the board

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

• Where the question came from (§5.4, p. 98). The chapter posed the four-point question at the start of the chords section and said outright that it would have to wait until the end of the chapter, because more machinery was needed first. That deferral is worth reminding the student of — it makes §5.8 the payoff of everything between.
• The given, as the chapter sets it (§5.8, p. 111). A segment AB; points C and D both on the same side of AB and neither on the line AB; and the angles ACB and ADB equal. To show: A, B, C and D lie on one circle. Note that the chapter states the not-on-the-line condition separately — a point on line AB subtends no angle there at all.
• Fig. 5.27 A (p. 111). The case where D is outside the circle. Circle through A, B and C: A on the left, B high on the right, C on the lower left, all on the circle. D sits at the bottom right, outside the circle. The segment from A to D crosses the circle at E, marked just above D. Segments drawn: the chord AB, the segments to C, the line A–E–D, and the segment from B to D, giving the small triangle B, E, D.
• Fig. 5.27 B (p. 111). The case where D is inside the circle. Same three points on the circle, but D now sits inside, and E is found by continuing AD until it meets the circle — so E is beyond D rather than before it, and lies on the circle at the lower right. The small triangle B, E, D is again present, but D and E have swapped roles.
• The argument, as data. In both figures C and E are on the circle and on the same side of the chord AB, so the angles ACB and AEB are equal — that is Equal angles in the same segment: the arc looks the same from every point beyond it being used. Suppose D is outside: then the angle AEB is exterior to triangle BED, so it strictly exceeds the angle ADB. Combining with the equality just noted and the given equality of the angles at C and D, the angle ACB comes out strictly greater than itself. Suppose D is inside: then the angle ADB is the exterior angle of that triangle instead, and the same chain runs the other way to the same impossibility. Both suppositions fail, so D is on the circle.
• A numerical dry run the explanation can show (not in the book; the chapter prints none). Chord AB with an on-circle reading of 50° on the relevant side, and D given with the angle ADB equal to 50°. Verified: if D were outside, its reading would have to be under 50°; if inside, over 50°. Neither is 50°, so D is on the circle. Presenting the contradiction as an arithmetic squeeze first, then as the formal exterior-angle argument, makes the logic land.
• Why the same-side condition is not decoration. Verified: take a chord AB whose far arc reads 50°. A point on the near arc reads 130°. Both points are on the circle, but their angles are not equal — so equality is not necessary for concyclicity when the points are on opposite sides. Conversely, put C on one side reading 50° and D on the other side reading 50°: those two are not concyclic with A and B, because a concyclic D on that side would have had to read 130°. That second case is the one Theorem 10 excludes by fiat.
• Exercise Set 5.6, Q2(iii) (p. 111). Inputs: X and Y not on the circle, with the angles AXB and AYB equal; is Y picked up by the circle drawn through A, B and X? Verified: yes provided X and Y sit to one side of AB, both of them — Theorem 10's own hypothesis, which this question's wording drops. Without it the answer is no: place X and Y symmetrically on opposite sides of AB, each seeing it at 120°, and the circle through A, B and X has Y for its centre rather than passing through it. Only the 90° case survives on opposite sides. This is the same caution as the Misconception bullet below, and the two must agree. The chapter places the question on the same page as the theorem, so the reader meets it as a puzzle and then as a result.
• Where the naming lands (§5.8, p. 112). Once four points are known to be concyclic, the chapter names the figure their vertices make: a cyclic quadrilateral, and it introduces 4-gon as an alternative word in the same sentence. Both words are printed in bold and both are the chapter's.

Figures to have open

• Figs. 5.27 A and B redrawn as a pair on one screen, with the small triangle B-E-D highlighted in each and the difference in E's position made obvious. These are the chapter's own figures (p. 111); printed side by side but with different vertex layouts, so the correspondence is hard to see.
• A single isolated triangle for section 7: B, E, D with the exterior angle marked and the inequality arrow drawn. Standard schematic; the chapter's version is prose on p. 112.
• A contradiction panel: the chain of equalities and one inequality laid out so that the same symbol appears at both ends. Standard schematic; not in the book, and it is what makes an argument by contradiction legible to a Class 9 student.
• A same-side / opposite-side comparison for section 10, with a chord and points on both arcs carrying their readings. Standard schematic; the chapter states the condition and draws no counter-example.
• 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 section opens on p. 111 with the definition of concyclic and Theorem 10; the argument runs from the foot of p. 111 through the first half of p. 112, where the cyclic quadrilateral is named.
• Figs. 5.27 A and 5.27 B, p. 111.
• Exercise Set 5.6, Q2, pp. 110–111 — Q2(iii) is this topic's question.
• Back-reference inside the chapter: the four-point question is posed and deferred at §5.4, p. 98. Theorem 1, which the proof depends on, is at §5.3, p. 96. The equal-angle property the proof uses is at §5.7.1, p. 109, and the direction of the off-circle inequalities is the same p. 112 argument seen from Off the circle the angle changes: points inside and outside compared (Fig. 5.25).
• Chapter Summary, p. 117 — the concyclicity bullet.