SSS: sides in proportion drag the angles into agreement

What to assume they know

• The two conditions polygons must both meet — the square-and-rhombus counter-example, which is exactly what makes this result surprising
• Turning it around: matching ratios force the line to be parallel — the converse of the Basic Proportionality Theorem
• AAA and AA: equal angles are enough on their own — the AAA criterion and its cut-and-copy proof
• SSS congruence of triangles from Class IX
• Comparing three fractions for equality, including with a surd in numerator and denominator
• The angle sum property of a triangle

What they should be able to do

• State the question SSS answers, and say why the answer is not obvious given the quadrilateral counter-example
• Carry out Activity 5 and verify that the three side ratios agree
• State the SSS criterion for similarity and distinguish it from SSS congruence
• Reconstruct the proof, naming the point at which the converse of the Basic Proportionality Theorem is used
• Explain why deriving BC = PQ is the step the proof turns on
• Explain the remark that, for triangles, checking one clause makes the other unnecessary
• Test three given side ratios and write the resulting similarity with the vertices in the right order
• Show that a pair of triangles fails SSS under every possible pairing, not just the one written down

Where it usually goes wrong

• "Sides and angles are independent, so proportional sides cannot force angles." True of quadrilaterals, false of triangles. The rigidity of the triangle is exactly what separates the two cases, and the chapter has just shown a rhombus failing where a triangle would succeed.
• "The ratio has to be less than 1." The proof is written that way only so the copy fits inside the larger triangle. Swap which triangle you call which and the same argument runs.
• "SSS similarity and SSS congruence are the same rule." Congruence wants three equal pairs; similarity wants three pairs in one ratio. Congruence is the case where that ratio is 1 — and note that the similarity proof calls on the congruence rule at its last step, so they are not even competitors.
• "Write the similarity alphabetically." Example 5 pairs ABC with RQP and question 1(ii) pairs ABC with QRP. Get the order from the ratios, never from the alphabet.
• "3.8 over 7.6 with surds in the other fraction cannot possibly match." The √3 cancels. Students abandon the check the moment a surd appears.
• "Two of the three ratios agreeing is enough." Item (iii) has two ratios landing on 0.5 once the sides are sorted, and it is still not similar.
• "If the written pairing fails, the triangles are not similar." Not yet proved — you have to rule out the other pairings, which is what sorting the sides does in one move.

Questions to check understanding

• State and prove the SSS similarity criterion with reasons at each step
• Given six side lengths, decide similarity and write the statement in the correct vertex order
• Given six side lengths that fail, show they fail under every correspondence
• Find an unknown angle by pairing a similarity with the angle sum, as in Example 5
• Explain the difference between the SSS similarity rule and the SSS congruence rule
• Prove a similarity in a lettered figure where the ratios come from medians or bisectors — Exercise 6.3 questions 10, 12, 14 and 16
• One-mark recall: which clause of the polygon definition may be skipped for triangles, and why

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.

• Activity 5 and Fig. 6.25 (p. 88, checked). Two triangles are drawn to order: ABC with AB = 3 cm, BC = 6 cm, CA = 8 cm, and DEF with DE = 4.5 cm, EF = 9 cm, FD = 12 cm. On the page A sits at the upper left of the first triangle with B below it and C to the right, and D sits at the top of the second with E below left and F to the right; the label 8 cm is set on AC and 12 cm on DF. Verified: 3/4.5 = 2/3, 6/9 = 2/3 and 8/12 = 2/3, so the three ratios agree exactly. Verified as a sanity check on the drawing: 3 + 6 = 9, which exceeds 8, so ABC really is a triangle, though a distinctly flat one — which is why it is drawn long and low on the page. The student is then asked to measure all six angles and finds the three pairs equal.
• Theorem 6.4 and Fig. 6.26 (pp. 88–89). Take triangles ABC and DEF whose three side ratios agree, the ratio taken smaller than 1 so the copy fits inside. Mark P on DE with DP = AB and Q on DF with DQ = AC, and join PQ. The given proportion rearranges into DP/PE = DQ/QF, and the converse of the Basic Proportionality Theorem then makes PQ parallel to EF — so ∠P = ∠E and ∠Q = ∠F. Those two equalities leave triangles DPQ and DEF equiangular, and it is Theorem 6.3, proved on the page before, that turns equiangularity into all three ratios: DP/DE = DQ/DF = PQ/EF. The Basic Proportionality Theorem cannot reach that third ratio; it speaks only about the two cut sides. But the original hypothesis already said DP/DE = DQ/DF = BC/EF. Comparing, PQ/EF = BC/EF, so BC = PQ. Now ABC and DPQ have all three sides equal, so they are congruent by SSS, and the angle equalities ∠A = ∠D, ∠B = ∠E, ∠C = ∠F follow. The step to dwell on: the proof spends its whole middle converting a proportion into the single equality BC = PQ, because that is the one thing congruence needs and proportion does not give.
• The Remark (p. 89). Neither clause alone settles similarity for polygons — the square and the rhombus are still standing. But Theorems 6.3 and 6.4 together mean that for triangles, verifying one clause makes the other automatic. Verified as a contrast worth showing: the rhombus of Fig. 6.7 (p. 78) has all four side ratios equal to 2 against its square and is still not similar to it; a triangle in the same position would be.
• Example 5 and Fig. 6.30 (pp. 91–92, checked). Triangle ABC carries AB = 3.8, BC = 6, CA = 3√3, with ∠A = 80° and ∠B = 60° marked on the drawing. Triangle PQR carries PQ = 12, QR = 7.6 and RP = 6√3, with P at the lower left, Q at the lower right and R at the top; no angle is marked on it. The question asks for ∠P. Verified: AB/RQ = 3.8/7.6 = ½, BC/QP = 6/12 = ½, and CA/PR = 3√3 ÷ 6√3 = ½ — the surd cancels, which is the only reason those particular lengths were chosen. So the similarity is ABC with RQP, in that order. Verified: ∠C = 180 − 80 − 60 = 40°, and ∠C pairs with ∠P, so ∠P = 40°. The pairing is the lesson here — the natural guess ABC with PQR is wrong, and a student who writes it gets ∠A's partner instead of ∠C's.
• Exercise 6.3 question 1 item (ii), Fig. 6.34 (p. 95, checked). Triangle ABC with AB = 2, BC = 2.5, CA = 3; triangle PQR with QR = 4, RP = 5, PQ = 6. Verified: 2/4 = 2.5/5 = 3/6 = ½, so ABC is similar to QRP, again in an order that is not the alphabetical one.
• Exercise 6.3 question 1 item (iii), Fig. 6.34 (p. 95, checked). Triangle LMP with LM = 2.7, MP = 2, PL = 3; triangle DEF with DE = 4, EF = 5, FD = 6. Verified in the written order: 2.7/4 = 0.675, 2/5 = 0.4, 3/6 = 0.5 — three different numbers. Verified for every other pairing: sorting each triangle's sides gives 2, 2.7, 3 against 4, 5, 6, and the ratios of the sorted lists are 2/4 = 0.5, 2.7/5 = 0.54 and 3/6 = 0.5 — still not all equal, and the smallest must go with the smallest under any similarity. So no correspondence works and the pair is not similar. That second check is what turns "the ratios I wrote down disagree" into a proof.
• Example 8 part (iii) (p. 94, checked). Within the median problem, the final part is settled by SSS similarity: once CM/RN is shown equal to BC/QR and to BM/QN, triangles CMB and RNQ are similar. Included here as the place a student meets SSS with letters rather than numbers.

Figures to have open

• Fig. 6.26 (pp. 88–89) redrawn so triangle DPQ inside DEF can be lifted out and laid over ABC. The same shown step by step asset as the AAA topic's Fig. 6.24, which is worth building once and reusing — the visual identity of the two constructions is the thesis.
• Activity 5's pair from Fig. 6.25 (p. 88), drawn to the printed lengths so the flatness of the 3, 6, 8 triangle is visible.
• Fig. 6.30's pair (p. 91) with all six lengths and the two angles printed on them, including the surds as printed.
• Items (ii) and (iii) of Fig. 6.34 (p. 95) with their six lengths each, for the closing comparison.
• A recall panel of the square-and-rhombus pair from Fig. 6.7 (p. 78) for section 2 and section 10. All standard schematics.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 6 "Triangles", §6.4, pp. 88–89: Activity 5, Theorem 6.4 with its proof, and the Remark that follows. Figures 6.25 and 6.26.
• Example 5, pp. 91–92, with Fig. 6.30; Example 8 part (iii), p. 94, with Fig. 6.33.
• Exercise 6.3 question 1 items (ii) and (iii), p. 95, with Fig. 6.34.
• Item 8 of the chapter's summary, p. 98.