A line drawn parallel to a side cuts the other two in matching ratios

What to assume they know

• The two conditions polygons must both meet — the two-clause definition of similarity for polygons
• Area of a triangle as half of base times height, and that either side may be taken as the base provided the matching perpendicular is used
• The Class IX result that triangles on one base with their opposite vertices on a line parallel to it have equal areas
• Dropping a perpendicular from a point to a line, and the notation for it
• Simplifying a ratio of two fractions that share a factor
• That a ratio equation may be read in either direction

What they should be able to do

• Perform the construction of Activity 2 and report the two ratios it produces
• Explain why a measured agreement in an activity is evidence and not proof
• State what the theorem claims about a triangle cut by a line parallel to one of its sides
• Identify, in the proof's figure, which two segments are joined and which two perpendiculars are dropped, and say what each is for
• Write the area of each of the four small triangles in terms of one of the four segments and one of the two perpendiculars
• Show that the ratio of the first pair of areas simplifies to a ratio of lengths on one side, and the second pair to a ratio of lengths on the other
• Identify the single step at which the parallel hypothesis is used
• Explain why the two ratios need not equal 1, and what it would take for them to

Where it usually goes wrong

• "The parallel line bisects the two sides." Only when it is the midline. Activity 2 splits them 3 to 2 and the theorem is perfectly happy.
• "The theorem says AD/AB = AE/EC." It does not. The two fractions must be built the same way on both sides — either both from the two pieces, AD/DB and AE/EC, or both from a piece and the whole, AD/AB and AE/AC. Mixing the two forms is the single commonest slip on this theorem.
• "DM and EN are the triangle's altitudes." They are not. DM runs from D perpendicular to AC and EN from E perpendicular to AB; neither starts at a vertex of ABC.
• "BDE and DEC are equal in area because they are congruent." They are generally not congruent at all — they can have quite different shapes. They are equal in area because they stand on one segment between one pair of parallels, which is a much weaker condition than congruence.
• "The area argument is a trick; there must be a way with lengths alone." Area is what converts a length ratio into something two different triangles can share. That is the idea worth taking away.
• "Measuring in Activity 2 proves it." Measurement can only ever say that it worked this time, to within the accuracy of a ruler.
• "The two points where the line cuts the sides could coincide." The statement rules that out by requiring two distinct points; a line through the apex meets the sides at one point and says nothing.

Questions to check understanding

• State the theorem and identify the hypothesis in a given figure
• Reproduce the proof, with the reason supplied at each step — this is a standard full-mark proof question on this chapter
• Given three of the four segments on the two cut sides, find the fourth
• Name the property of areas used in the proof and say where it is used
• Prove with this theorem that a line through one side's midpoint, drawn parallel to a second side, cuts the remaining side into two equal pieces — Exercise 6.2 question 7
• Explain why the perpendiculars can be dropped anywhere the construction requires without affecting the result

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 2 and Fig. 6.9 (p. 79, checked). Draw an angle XAY. Along the arm AX step off five equal lengths, lettering the marks in order P, Q, D, R, B, so that each of AP, PQ, QD, DR, RB measures the same. Through B draw any line meeting the other arm AY at C. Through D run a second line, this one parallel to BC; it meets AC at E. Verified: counting the equal steps, AD spans three of them and DB spans two, so AD/DB = 3/2 by construction and not by measurement. The activity then has the student measure AE and EC and find AE/EC comes out at 3/2 as well. In the printed drawing A sits at the left with AY running up to the right and AX down to the right, so the five stepped marks lie on the lower arm and E with C on the upper one.
• Fig. 6.10 and the proof (p. 80, checked). Triangle ABC with A at the apex, B at lower left and C at lower right; D on AB and E on AC with DE drawn across. For the proof, BE and CD are drawn in — they appear as dashed lines on the page — and two perpendiculars are added: DM at right angles to AC, and EN at right angles to AB. The four areas the proof needs are then
  • the triangle ADE, taken on base AD with height EN: ½ × AD × EN
  • the triangle BDE, taken on base DB with the same height EN: ½ × DB × EN
  • the triangle ADE again, this time on base AE with height DM: ½ × AE × DM
  • the triangle DEC, on base EC with the same height DM: ½ × EC × DM Verified: dividing the first by the second leaves AD/DB, and dividing the third by the fourth leaves AE/EC — the ½ and the perpendicular cancel in each case. Note when explaining it that the triangle ADE is measured twice, on two different bases, and this is the move the whole proof turns on.
• The one use of parallelism (p. 80). BDE and DEC both stand on the segment DE, and their remaining vertices B and C both lie on BC, which is parallel to DE. So the two have equal areas. With the two denominators equal, the two fractions AD/DB and AE/EC are both equal to the same thing, and therefore to each other.
• A concrete instance for showing it. Take the activity's own numbers, AD = 3 units and DB = 2 units, and let the perpendicular EN measure 4 units. Verified: the area of ADE is then 6 square units and the area of BDE is 4, and 6/4 reduces to 3/2. Change EN to 10 and the two areas become 15 and 10 — the same 3/2.
• What the theorem does not promise. Verified: nothing forces AD/DB to be 1. In Activity 2 it is 3/2, and the parallel line is nowhere near the middle of AB. The bisecting case is one instance among infinitely many, and it is set as questions 7 and 8 of Exercise 6.2 (p. 85) precisely because it is a special case rather than the content.
• Thales (p. 79). Named with the dates 640 to 546 B.C. and shown in a captioned portrait panel on the page. The chapter attributes to him the claim that in two equiangular triangles the ratio of any two matched sides is always the same, and says it is believed he reached it using this theorem.

Figures to have open

• Fig. 6.10 (p. 80) redrawn as the working diagram of the whole topic: triangle ABC, D on AB, E on AC, DE drawn, BE and CD added, DM perpendicular to AC and EN perpendicular to AB. Every later section builds on this one figure, so it must be drawn once and kept in view. Redraw as a schematic; do not trace the printed art.
• The Activity 2 construction of Fig. 6.9 (p. 79): the angle XAY, five equal marks on one arm, the line through B, and the parallel through D. Standard schematic.
• A shaded-area panel: the segment DE with two triangles standing on it, their apexes at B and at C on a line parallel to DE, shaded to equal areas. Standard schematic and the most important single visual in the topic.
• A portrait card for Thales with the dates 640 to 546 B.C. The textbook's own portrait need not be reproduced.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 6 "Triangles", §6.3 Similarity of Triangles, pp. 79–81, covering the restatement of the two clauses for triangles, the Thales panel, Activity 2, and Theorem 6.1 with its proof. Figures 6.9 and 6.10.
• The result is item 4 of the chapter's summary, p. 97.
• Exercise 6.2 questions 7 and 8 (p. 85) apply it to the midpoint case; those belong to the next topic's exercise sweep and are named here only as the place the special case is handled.