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Chapter 1 · Geometric Twins

ASA and AAS: two angles are enough to fix the third

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10 min.

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What the video covers, what to say before it, where a class usually goes wrong, and what to set afterwards. An account is free, and it opens every chapter of every book.

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What to assume they know

What they should be able to do

  • State the ASA condition and identify the included side for a stated pair of angles
  • Construct a triangle from one side and the two angles at its ends
  • Recognise a crossing at a common midpoint as an SAS situation, and say which three equalities it supplies
  • Use vertically opposite angles as an equality that costs no measurement
  • Read off the correspondence in a crossing figure, and deduce that two segments are equal
  • Compute the third angle of a triangle from the other two
  • Convert an AAS set of givens into an ASA set, and say why that is legitimate
  • Explain why AAS is not an independent condition

Where it usually goes wrong

  • "ASA and AAS are two different rules to memorise." They are one rule and one consequence of it. The chapter derives AAS in six printed lines using nothing but the angle sum. A student who understands the derivation cannot forget the rule.
  • "If the given side is not between the angles, you are stuck." That is the SSA reflex being applied where it does not belong. With two angles given, the third is free, so no configuration of one side and two angles is ever underdetermined. Contrast this explicitly with the SSA case on p.12.
  • "You need to measure the third angle." You compute it. The chapter's subtraction from 180° is arithmetic on paper, not a protractor reading, and it is exact in a way a measurement is not.
  • "Two angles determine the triangle." They determine its shape only. The three same-angle triangles on p.10 already showed that; the side is what fixes the size, and AAS still needs one.
  • "This figure is under the ASA heading, so it must be an ASA problem." The crossing figure on pp.12–13 sits under that subheading and is settled by SAS. The book says so; a script that skims the headings will get it wrong.
  • "Vertically opposite angles are equal only when the figure looks symmetrical." They are equal at every crossing. Here that fact is worth a measurement, and it is the reason two midpoints are enough.
  • "Congruence is the end of the problem." In the crossing figure congruence is the middle of it. The question asked how AB compares with CD, and the congruence is only the instrument that answers it.

Questions to check understanding

  • Given a side and the two angles at its ends, construct the triangle
  • Identify the included side for a stated pair of angles
  • Given two angles of a triangle, find the third
  • Given an AAS set of equalities, show that ASA applies and express the congruence
  • Given two segments crossing at a common midpoint, prove that a pair of opposite sides are equal (the shape of the worked figure, Part II, pp.12–13)
  • Use parallel lines and alternate angles to obtain the angle equalities a congruence needs (the shape of Figure it Out question 2, Part II, pp.13–14)
  • Show that two angles are equal by first showing the triangles containing them are congruent (the shape of Figure it Out question 3, Part II, p.14)
  • Sort measurement sets by which condition they satisfy — case (d) of Figure it Out question 2 (Part II, p.20) is an AAS set

Examples worth working on the board

  • The ASA case (Part II, §1.2, p.12). Inputs: BC and YZ are both 5 cm; ∠B and ∠Y are both 50°; ∠C and ∠Z are both 30°. The equal side runs between the two equal angles. As with SAS, the chapter has the reader construct rather than printing the finished triangles.
  • The crossing figure (Part II, §1.2, p.12). Checked against the printed page. Four points and one crossing: A at the upper left, C at the upper right, B at the lower left, D at the lower right, with segments AD and BC drawn so they cross at O. Tick marks confirm AO = OD (single) and BO = OC (double). Inputs: O is the shared midpoint of both those segments. Everything else is deduced.
  • The three equalities (Part II, §1.2, pp.12–13). Two come from the midpoints; the third is that ∠AOB and ∠DOC are vertically opposite. Note: the chapter places this worked figure inside the ASA subheading, but the condition that applies to it is SAS, and the chapter says so plainly on p.13. Do not let the placement mislead the script.
  • What congruence hands back (Part II, §1.2, p.13). Once ΔAOB ≅ ΔDOC is established with the correspondence A↔D, O↔O, B↔C, the segments AB and DC turn out to be corresponding sides, so they are equal. That answers the question the figure was posed with, and it is a length nobody measured.
  • Fig. 1.2 (Part II, §1.2, p.15). Checked against the printed page. Two tall thin triangles side by side, each with the apex at the top: A above, B lower left, C lower right; X above, Y lower left, Z lower right. Marked: 35° at A and at X, 75° at C and at Z, and the base tagged 4 cm in both. The caption "Fig.1.2" sits beneath the pair. The equal side is BC, which does not lie between the two equal angles — that is what makes this AAS.
  • The angle-sum step (Part II, §1.2, p.15). Inputs: 35° and 75° in each triangle, and a total of 180°. The chapter then reprints the same two triangles with the newly-found angle written in at B and at Y — a second figure that differs from the first only by that one label. Showing both is the whole point.
  • The rewritten problem (Part II, §1.2, p.15). Once the third angle is in hand, the givens are ∠B = ∠Y, BC = YZ and ∠C = ∠Z, and the side now does lie between the two angles. The chapter lists these three lines and closes.
  • Figure it Out, question 2 (Part II, §1.2, pp.13–14). Checked against the printed page. D upper left, C upper right, A lower left, B lower right, with segments DB and AC crossing at O, and arrow-plus-tick marks on DC and on AB showing they are both parallel and equal. Inputs: CD parallel to AB, CD = AB, and the printed hint that alternate angles are equal.
  • Figure it Out, question 3 (Part II, §1.2, p.14). Checked against the printed page. A at the top, D at the bottom, B and C at left and right with BC drawn between them; arcs mark equal angles at B and at C on both sides of BC. Inputs: ∠ABC = ∠DBC, ∠ACB = ∠DCB, and BC common to both triangles. A clean ASA item.
  • Figure it Out, question 4 (Part II, §1.2, p.14). Checked against the printed page. A upper left, D upper right, B lower left, C lower right, with AC and DB drawn crossing between them. Inputs: ∠ABD = ∠DCA and ∠ACB = ∠DBC. The task is to list the equal parts, not to name one condition.

Figures to have open

  • A triangle built from one side and the two angles at its ends, able to be shown moving so the two rays sweep out and meet. Standard schematic.
  • The crossing figure with a common midpoint, lettered A, C above and B, D below with O at the crossing, tick marks on both pairs. Standard schematic; keep the printed lettering so the correspondence A↔D, B↔C reads correctly.
  • Fig. 1.2's pair of triangles, redrawn, with the 35°, 75° and 4 cm marks in the printed positions, and a second state in which the computed angle appears at B and at Y. The two-state version is what carries section 8.
  • A small tree of the five sufficient conditions, for section 10.
  • No photograph or data table from the textbook is needed.

Where this sits in the book

  • NCERT Ganita Prakash, Class 7, Part II, printed Chapter 1 "Geometric Twins", §1.2 "Congruence of Triangles", pp.12–13 — the unnumbered subheading "Two Angles and the Included Side", the naming of ASA, and the crossing figure with the common midpoint (p.12); the vertically opposite angles, the SAS verdict and the subheading "What are the Corresponding Vertices?" (p.13)
  • Same part, same chapter, §1.2, "Measuring Two Angles and a Non-Included Side", Fig. 1.2 and the angle-sum step, pp.14–16
  • Same part, same chapter, §1.2, "Figure it Out", questions 2, 3 and 4, pp.13–14
  • Same part, same chapter, SUMMARY, p.22, the ASA and AAS bullet
  • Backward pointer: SAS, and why the angle has to be the included one for the SSA contrast this topic leans on
  • Forward pointer: Angles opposite equal sides are equal, where a congruence proved by these conditions is used to establish a property of a single triangle

The book

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