Turning it around: matching ratios force the line to be parallel

What to assume they know

• A line drawn parallel to a side cuts the other two in matching ratios — the forward theorem and its area proof
• What a converse of a statement is, and that its truth is a separate question
• Manipulating a proportion: inverting both sides, and adding 1 to both sides
• Corresponding angles and alternate angles at a transversal, from Class IX
• That in a triangle, sides opposite equal angles are equal
• That two lines parallel to a third line are parallel to each other

What they should be able to do

• State the converse of the Basic Proportionality Theorem and distinguish its hypothesis from the forward theorem's
• Explain why a true statement's converse is not automatically true, with an example from outside geometry if helpful
• Report the four ratios produced by Activity 3 and say what each accompanying parallel line demonstrates
• Reconstruct the converse's proof: assume failure, construct the parallel line, apply the forward theorem, and derive the collision
• Explain how adding 1 to both sides of the derived proportion forces two points to coincide
• Convert between the two-piece form of a proportion and the piece-to-whole form
• Decide, from four given lengths, whether a segment drawn across a triangle runs parallel to the remaining side
• Use the converse to obtain parallelism and then chase angles to a conclusion about the triangle

Where it usually goes wrong

• "If a statement is true its converse is true." Not in general. The chapter spends an activity and a proof establishing that this particular converse holds, which would be wasted effort if converses came free.
• "The converse is proved by running the area proof backwards." It is not. The area computation gives you a ratio from parallelism and has no reverse gear; the converse needs the assume-and-collide argument instead.
• "Proof by contradiction is a dodge." Here it is doing precise work: it gives you a second line to compare against, and the whole proof is the comparison.
• "AD/DB = AE/EC and AD/AB = AE/AC are different theorems." They are the same fact in two dresses, and Example 1 is the changing room. Students who never see the conversion treat them as separate results to memorise.
• "In question 2(iii) you can just divide the two given lengths." You can, as it happens, because the piece-to-whole form is legitimate — but only if you know Example 1. A student who reaches for PE/EQ must subtract first.
• "A drawing that looks parallel is parallel." The whole point of question 2 is that three plausible-looking figures give two verdicts.
• "The converse needs the ratio to be 1." Activity 3 runs it at 1/4, 2/3, 3/2 and 4/1, and it holds at every one.

Questions to check understanding

• State the converse and prove it, giving the reason at each step
• Given four lengths in a triangle, decide whether a drawn segment runs parallel to the remaining side, with justification
• Given three lengths and a parallel segment, compute the fourth
• Prove the midpoint theorem and its converse using the two theorems in turn — Exercise 6.2 questions 7 and 8
• Trapezium diagonal questions in both directions — Exercise 6.2 questions 9 and 10, a standard board pairing
• Chain the converse with an angle property to prove a triangle isosceles, in the manner of Example 3
• Convert a proportion from the two-piece form to the piece-to-whole form and back

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 3 and Fig. 6.11 (p. 81, checked). An angle XAY is drawn. On the ray AX five points are marked, B₁, B₂, B₃, B₄ and B, cutting it into five equal pieces; on AY five points C₁, C₂, C₃, C₄ and C do the same. Then B₁C₁ and BC are joined, and afterwards B₂C₂, B₃C₃ and B₄C₄. The page prints four ratio statements, each with the same value on both arms: 1/4 for the first pair of points, 2/3 for the second, 3/2 for the third, 4/1 for the fourth — and each of the four joining segments comes out parallel to BC. Verified: the four values are just k/(5 − k) for k running from 1 to 4. Note what that does and does not give you: the set is closed under taking reciprocals, since 1/4 pairs with 4/1 and 2/3 with 3/2, so it is balanced about 1 in the sense that matters for ratios — what it is not is evenly spaced, and two of the four land above 1 while two land below. On the printed drawing A sits at the top right, with B₁ to B₄ and B running down to the left towards X, and C₁ to C₄ and C running down to the right towards Y.
• Theorem 6.2 and Fig. 6.12 (p. 82). The proof takes a segment DE inside triangle ABC with AD/DB = AE/EC and supposes it is not parallel to BC. Then a second segment DE′ is drawn from D, parallel to BC, meeting AC at E′. The forward theorem gives AD/DB = AE′/E′C, so AE/EC = AE′/E′C. Adding 1 to both sides turns each fraction into a piece-to-whole ratio with the same denominator structure, and the two points E and E′ must be the same point — which contradicts the supposition that DE was not the parallel one. Note: the printed proof leaves three steps as Why? prompts, so the explanation is filling in a proof the book deliberately leaves partly open.
• The add-1 move, spelled out. Verified: from AE/EC = AE′/E′C, add 1 to each side to get (AE + EC)/EC = (AE′ + E′C)/E′C, that is AC/EC = AC/E′C, so EC = E′C, so E and E′ are the same point of AC. Worth working through slowly — it is the step every student skips.
• Example 1 and Fig. 6.13 (p. 82). D on AB, E on AC, DE parallel to BC. The target is AD/AB = AE/AC. Verified: the forward theorem gives AD/DB = AE/EC; invert both sides to DB/AD = EC/AE; add 1 to reach AB/AD = AC/AE; invert once more. This is the identity that lets a student move freely between the two-piece form and the piece-to-whole form.
• Example 2 and Figs. 6.14, 6.15 (p. 83, checked). A trapezium ABCD with AB parallel to DC; E on the non-parallel side AD and F on the non-parallel side BC, with EF parallel to AB. The target is AE/ED = BF/FC. The construction joins the diagonal AC, which meets EF at G. Since EF is parallel to AB and AB to DC, EF is parallel to DC. Inside triangle ADC the segment EG is parallel to DC, giving AE/ED = AG/GC; inside triangle CAB the segment GF is parallel to AB, giving CG/AG = CF/BF, which inverts to AG/GC = BF/FC. The two right-hand sides meet. The idea to carry: one diagonal converts a quadrilateral problem into two triangle problems, and the diagonal's own two pieces are the bridge.
• Example 3 and Fig. 6.16 (pp. 83–84, checked). In a triangle PQR, S lies on PQ and T on PR with PS/SQ = PT/TR, and additionally ∠PST = ∠PRQ. The converse gives ST parallel to QR; corresponding angles then give ∠PST = ∠PQR; combining with the given equality, ∠PRQ = ∠PQR; so PQ = PR and the triangle is isosceles. Note the shape of the argument — the converse is used to manufacture a parallel line whose only purpose is to license an angle equality.
• Exercise 6.2 question 1, Fig. 6.17 (p. 84, checked). Two triangles with DE parallel to BC. In (i): AD = 1.5 cm, DB = 3 cm, AE = 1 cm, and EC is to be found. In (ii): DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm, and AD is to be found. Verified: in (i), 1.5/3 = 1/EC gives EC = 2 cm. In (ii), AD/7.2 = 1.8/5.4 = 1/3 gives AD = 2.4 cm.
• Exercise 6.2 question 2 (p. 84, no figure). Three cases in a triangle PQR with E on PQ and F on PR, asking each time whether EF runs parallel to QR. The measurements, all in centimetres:

| case | on side PQ | on side PR | |---|---|---| | (i) | PE is 3.9, with EQ 3 | PF is 3.6, with FR 2.4 | | (ii) | PE is 4, with QE 4.5 | PF is 8, with RF 9 | | (iii) | the whole side PQ is 1.28, and PE is 0.18 | the whole side PR is 2.56, and PF is 0.36 |

Verified: (i) 3.9/3 = 1.3 against 3.6/2.4 = 1.5 — unequal, so not parallel. (ii) 4/4.5 = 8/9 exactly, so parallel. (iii) the whole sides are given rather than the pieces, so first EQ = 1.28 − 0.18 = 1.10 and FR = 2.56 − 0.36 = 2.20; then 0.18/1.10 = 9/55 and 0.36/2.20 = 9/55 — equal, so parallel. Item (iii) is a deliberate change of form and rewards the identity from Example 1.

• The rest of Exercise 6.2 (pp. 84–85, checked). Question 3 uses Fig. 6.18 with LM parallel to CB and LN parallel to CD, target AM/AB = AN/AD. Question 4 uses Fig. 6.19 with DE parallel to AC and DF parallel to AE, target BF/FE = BE/EC. Question 5 uses Fig. 6.20, where P is the apex over a base QR, O lies inside, and DE parallel to OQ and DF parallel to OR give EF parallel to QR. Question 6 uses Fig. 6.21, with A, B and C on OP, OQ and OR, and AB parallel to PQ and AC parallel to PR giving BC parallel to QR. Questions 7 and 8 are the two halves of the midpoint theorem, one for each of the two theorems. Questions 9 and 10 are a matched pair on a trapezium's diagonals: 9 assumes the trapezium and asks for the ratio equality AO/BO = CO/DO, and 10 assumes that same ratio equality and asks for the trapezium.
• The forward-and-converse pairing to point out. Verified as a pattern: questions 7 and 8 are one statement and its converse, and so are questions 9 and 10. The exercise set is built to make the student notice which theorem each direction needs.

Figures to have open

• Fig. 6.12 (p. 82) redrawn with E′ drawn away from E, so that the contradiction is visible before it is resolved. If the explanation draws them coincident from the start the proof has nothing to do. This is the topic's single most important figure.
• The Activity 3 arms of Fig. 6.11 (p. 81), with all five marks on each arm and the four joining segments capable of being switched on one at a time.
• The trapezium of Figs. 6.14 and 6.15 (p. 83) with the diagonal AC and the point G, drawn so the two triangles ADC and CAB can be lifted out separately.
• The two exercise triangles of Fig. 6.17 (p. 84) carrying their four printed lengths, for the closing section.
• All standard schematics; nothing here needs to be reproduced from the printed page.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 6 "Triangles", §6.3, pp. 81–84: Activity 3, Theorem 6.2 with its partial proof, and Examples 1, 2 and 3. Figures 6.11 to 6.16.
• Exercise 6.2, pp. 84–85, questions 1 to 10, with Figures 6.17 to 6.21.
• The result is item 5 of the chapter's summary, p. 97.
• The chapter points to Appendix 1 of the book for what a converse is — a deliberate cross-reference to pp. 218–238, outside this chapter's own folios.