AAA and AA: equal angles are enough on their own

What to assume they know

• The two conditions polygons must both meet — that for polygons neither clause implies the other
• A line drawn parallel to a side cuts the other two in matching ratios — the Basic Proportionality Theorem and its proof
• Turning it around: matching ratios force the line to be parallel — the converse, and the piece-to-whole conversion
• SAS congruence of triangles from Class IX, and the symbol for congruence
• The angle sum property of a triangle
• Corresponding, alternate and vertically opposite angles at intersecting lines
• Converting a length from centimetres to metres before forming a ratio

What they should be able to do

• Write a similarity statement for two triangles with the vertices in the correct order, and say which reorderings are still valid
• Carry out Activity 4 and report the ratio the three side pairs share
• State the AAA criterion, and explain why it is a theorem rather than a definition
• Reconstruct the proof: the cut, the congruence, the parallel segment, and the application of the Basic Proportionality Theorem
• Explain why the third angle is redundant, and restate the criterion as AA
• Apply AA to a figure in which the equal angles come from a transversal and a vertex crossing
• Set up and solve an indirect-measurement problem in which two right angles and a shared angle give the similarity
• Explain why equal angles say nothing whatever about the size of either triangle

Where it usually goes wrong

• "AAA works for congruence too." It does not, and the whole chapter exists because it does not. Equal angles fix shape and leave size entirely open.
• "Equal angles means equal sides." They mean proportional sides. The ratio can be anything at all, including a ratio nowhere near 1.
• "The order of letters is just labelling." It is the claim. Writing a similarity with A paired to the wrong vertex asserts something false, even if the two triangles genuinely are similar under some other pairing.
• "You have to check all three angles." Two suffice, every time, because the third has no freedom left.
• "In Example 7 the shadow is BE." BE is the whole ground distance from the post to the shadow's tip. The shadow itself is DE, the part beyond the girl.
• "90 and 3.6 can go straight into a ratio." One is in centimetres. This is the most reliable source of a wrong answer on Example 7.
• "The construction proves it by drawing a picture." The construction is legitimate because DP and DQ can always be marked off — DE and DF are longer than AB and AC when the smaller triangle really is smaller — and because SAS congruence, proved in Class IX, does the work at the joint.

Questions to check understanding

• State and prove the AAA criterion with reasons — a standard long-answer question
• Given a figure with two angle pairs marked, name the similar triangles in the correct symbolic order and cite the criterion
• Find an unknown angle by combining a similarity with the angle sum
• Indirect measurement: a pole, a tower and two shadows — Exercise 6.3 question 15, and the shadow-length problem of Example 7
• Explain why the third angle need not be checked
• Spot the error in a similarity statement whose vertex order is wrong
• Prove a similarity in a figure with altitudes or a transversal, where the angles have to be found before the criterion can be applied — Exercise 6.3 questions 5, 6, 7, 8 and 9

Examples worth working on the board

Values marked verified are worked out here on data printed inside pp. 73–98. The chapter prints no answers.

• Fig. 6.22 and the notation (p. 85, checked). Triangle ABC drawn large and triangle DEF drawn small. Verified by opening the page: the equal angles are indicated by arc marks drawn inside the artwork — one arc, two arcs, three arcs at the matched vertices — and no degree value is printed anywhere on either triangle. The next page, p. 86, introduces the tilde symbol for similarity — recalling the congruence symbol beside it — and carries the tinted, ruled box warning that the vertex order is part of the claim. Anyone sent to p. 85 for that box will not find it there. For these two triangles a statement pairing A with E, or A with F, is wrong, whereas reordering both triangles together — pairing B with E, A with D, C with F — is fine.
• Activity 4 and Fig. 6.23 (p. 86). Draw two segments of different lengths, BC = 3 cm and EF = 5 cm. At B and at C construct angles of 60° and 40°; at E and at F construct 60° and 40° likewise. The two pairs of rays meet at A and at D. Verified: the third angles are then 180 − 60 − 40 = 80° at both A and D, so the two triangles are equiangular. The book reports BC/EF = 3/5 = 0.6 and that measuring gives AB/DE and CA/FD equal to 0.6 as well, or nearly so where the ruler falls short. Verified as a consequence: if the ratio really is 0.6 then DE = AB/0.6 and FD = CA/0.6, so an explanation that draws AB = 2.4 cm must draw DE = 4 cm.
• Theorem 6.3 and the proof, Fig. 6.24 (p. 87). Take triangles ABC and DEF with all three angle pairs equal. On DE mark P so that DP = AB, on DF mark Q so that DQ = AC, and join PQ. Then ABC and DPQ agree in two sides and the angle between them, so they are congruent. Hence ∠B = ∠P, and since ∠B was equal to ∠E we get ∠P = ∠E — which are corresponding angles for PQ and EF cut by DE, so PQ is parallel to EF. The Basic Proportionality Theorem then gives DP/PE = DQ/QF, and the piece-to-whole conversion turns that into DP/DE = DQ/DF, that is AB/DE = AC/DF. Repeating on the other pair of sides finishes the ratio chain. Note: the printed proof marks four steps with Why? or How? prompts and leaves them open, and the piece-to-whole conversion is one of the things it does not spell out.
• The AA remark (p. 88). If two angle pairs agree, the third is forced by the angle sum, so the three-angle criterion is really a two-angle criterion. Verified: with 60° and 40° given, the third is 80° with no freedom at all.
• Example 4 and Fig. 6.29 (p. 91, checked). Segments PQ and RS are parallel; P is joined to S and Q to R, and those two crossing segments meet at O. Alternate angles give ∠P = ∠S and ∠Q = ∠R, and the angles at O are vertically opposite, so triangle POQ is similar to triangle SOR. Worth pointing out: the book cites the three-angle criterion here although two of the equalities would already have settled it.
• Example 7 and Fig. 6.32 (pp. 92–93, checked). Inputs: a girl 90 cm tall walks away from the foot of a lamp-post at 1.2 m/s; the lamp sits 3.6 m above the ground; the question is the length of her shadow after 4 seconds. In the figure AB is the post with A at the lamp, CD is the girl with D at her feet, and E is the shadow's tip, with B, D and E on one ground line. Verified: the distance walked is 1.2 × 4 = 4.8 m, so BD = 4.8 m. The post and the girl both stand vertical, so the angles at B and at D are right angles, and the angle at E is shared — AA gives triangle ABE similar to triangle CDE. Writing DE as x, the proportion BE/DE = AB/CD becomes (4.8 + x)/x = 3.6/0.9 = 4, so 4.8 + x = 4x, so 3x = 4.8 and x = 1.6 m. Note the unit trap: the girl's height must be converted, 90 cm = 0.9 m, before it meets 3.6 m in a ratio.
• Exercise 6.3 question 1, items (i) and (vi), Fig. 6.34 (pp. 94–95, checked). Item (i): triangle ABC with ∠A = 60°, ∠B = 80°, ∠C = 40°, and triangle PQR with ∠P = 60°, ∠Q = 80°, ∠R = 40°. Verified: all three pairs agree in that order, so ABC is similar to PQR. Item (vi): triangle DEF with ∠D = 70° and ∠E = 80° marked, and triangle PQR with ∠Q = 80° and ∠R = 30° marked. Verified: the missing angles are ∠F = 180 − 70 − 80 = 30° and ∠P = 180 − 80 − 30 = 70°, so D pairs with P, E with Q and F with R, and DEF is similar to PQR by the two-angle criterion. Item (vi) is the one that rewards computing a third angle before pairing.
• Exercise 6.3 question 15 (p. 97, checked). A vertical pole 6 m long casts a 4 m shadow; at the same moment a tower casts a 28 m shadow. Verified: the sun's rays make one angle for both, and both objects stand vertical, so AA applies and the height is 6 × 28 ÷ 4 = 42 m. This is the question the chapter's introduction was pointing at when it spoke of measuring what a tape cannot reach.

Figures to have open

• Fig. 6.24 (p. 87) redrawn so the cut-down copy can be shown moving: the larger triangle DEF, with a triangle DPQ inside it that can be lifted out and overlaid on ABC. Everything in sections 7 to 9 hangs on the student seeing that overlay.
• Fig. 6.22 (p. 85) with the arc marks, for the notation section. Arcs only.
• Fig. 6.29 (p. 91): two parallel segments with two crossing joins, so the alternate-angle pairs and the vertical pair can be lit separately.
• The lamp-post scene of Fig. 6.32 (p. 92) redrawn to scale: post 3.6 m, girl 0.9 m, ground distance 4.8 m, shadow x, all on one horizontal line. Drawing it roughly to scale makes the answer of 1.6 m look right before it is computed.
• A sun-ray diagram for the pole-and-tower question: 6 m with a 4 m shadow beside a tower with a 28 m shadow, the rays drawn parallel. Standard schematic.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 6 "Triangles", §6.4 Criteria for Similarity of Triangles, pp. 85–88: the restatement of the two clauses, the tilde notation and the boxed warning about vertex order, Activity 4, Theorem 6.3 with its proof, and the AA remark. Figures 6.22, 6.23, 6.24.
• Example 4, p. 91, with Fig. 6.29; Example 7, pp. 92–93, with Fig. 6.32.
• Exercise 6.3, pp. 94–97: question 1 items (i) and (vi) with Fig. 6.34, and question 15.
• Items 6 and 7 of the chapter's summary, p. 97.