SAS: one angle plus the two sides that enclose it

What to assume they know

• AAA and AA: equal angles are enough on their own — the AAA and AA criteria and the cut-and-copy proof
• SSS: sides in proportion drag the angles into agreement — the SSS criterion, and that one clause now implies the other for triangles
• SAS congruence from Class IX, and the meaning of an included angle there
• Vertically opposite angles at a crossing point
• Rearranging a product of two lengths equalling another product into a proportion
• What a median of a triangle is, and that it meets a side at its midpoint

What they should be able to do

• State the SAS criterion for similarity, with the enclosure condition stated explicitly
• Carry out Activity 6 and verify the two side ratios agree
• Reconstruct the proof, and say why the equal angle is what makes the congruence step available
• Rearrange an equality of two products into the proportion a similarity criterion can use
• Recognise a vertically opposite pair as the supplier of the enclosed angle
• Apply the criterion to a figure whose ratios come from medians rather than from printed lengths
• Decide whether a given pair of triangles satisfies SAS, and if not, say exactly which condition fails
• State the additional criterion the chapter offers for right triangles after the summary

Where it usually goes wrong

• "Any equal angle plus two proportional sides will do." Item (v) of the exercise is the refutation, and it is worth walking through slowly: both ratios are a half, both marked angles are 80°, and the pair still cannot be called similar.
• "Included just means the angle is in the triangle somewhere." It means the angle sits at the vertex where the two named sides meet. Nothing weaker.
• "OA · OB = OC · OD gives OA/OB = OC/OD." It does not. It gives OA/OC = OD/OB. Getting this wrong pairs the wrong triangles and the vertically opposite angle then sits in the wrong place.
• "The similarity in Example 6 can be written AOD with BOC." The pairing is forced by the proportion: A with C, O with O, D with B. Reversing the last two letters asserts something the working does not support.
• "Medians shrink the ratio." Both terms are halved, so the ratio is untouched — which is exactly why a median may stand in for a side.
• "SAS needs the angle to be a right angle." It does not. RHS is a separate and additional rule, and it names the hypotenuse specifically.
• "There must be an SSA criterion as well." The chapter gives none. What it does give, after the summary, is RHS — which is the one situation where naming a side beyond the enclosing pair works, and it works because the right angle removes the ambiguity.

Questions to check understanding

• State and prove the SAS similarity criterion with reasons
• Given two side ratios and one angle, decide whether the criterion applies, and if not say which condition fails — the standard trap is an angle that is not enclosed
• Turn a product equality into a proportion and identify the similar triangles, as in Example 6
• Median problems: show two triangles similar given sides and a median in proportion — Exercise 6.3 questions 12, 14 and 16
• Prove a ratio of medians equals the ratio of sides — Example 8 part (ii) and Exercise 6.3 question 16
• Angle-bisector problems in the same shape — Exercise 6.3 question 10
• One-mark recall on the RHS criterion for right triangles

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 6 and Fig. 6.27 (pp. 89–90, checked). Two triangles are drawn to order. In ABC, A sits at the top with AB = 2 cm running down to B and AC = 4 cm running across to C, and the angle at A is 50°. In DEF, D sits at the top with DE = 3 cm and DF = 6 cm, and the angle at D is 50°. Verified: AB/DE = 2/3 and AC/DF = 4/6 = 2/3, so the two ratios agree, and in each triangle the 50° is the angle between exactly those two sides. The student then measures the remaining four angles and finds them equal in pairs — at which point the book closes the activity by invoking the angle criterion rather than proving anything new, which is worth saying aloud in the explanation.
• Theorem 6.5 and Fig. 6.28 (p. 90, checked). Two triangles with AB/DE = AC/DF, the ratio taken below 1, and ∠A equal to ∠D. Mark P on DE with DP = AB and Q on DF with DQ = AC, and join PQ — on the printed page PQ is drawn as a dashed segment inside the larger triangle. The given proportion makes PQ parallel to EF, and DP = AB together with DQ = AC and ∠D = ∠A make triangles ABC and DPQ congruent by SAS. It is that congruence, not the parallel, that hands over ∠B = ∠P and ∠C = ∠Q. What the parallel supplies is the other pair, ∠P = ∠E and ∠Q = ∠F, and only by chaining the two do you reach ∠B = ∠E and ∠C = ∠F; with ∠A = ∠D given, Theorem 6.3 then delivers the similarity of ABC with DEF. Skip the parallel's contribution and the argument never mentions E or F at all. The point to make: the cut-and-copy construction is the very same one used for both earlier criteria. What differs is which fact yields the parallel — here and in Theorem 6.4 the given ratios do, through the converse of the Basic Proportionality Theorem, whereas Theorem 6.3 gets it out of a congruence it has already established.
• Example 6 and Fig. 6.31 (p. 92, checked). Two segments AB and CD cross at a point O — on the page A sits upper left, D lower left, C upper right and B lower right, so the two segments genuinely cross rather than merely touch. The hypothesis is the product equality OA · OB = OC · OD. Verified: dividing both sides by OC · OB rearranges it to OA/OC = OD/OB — note carefully which letters end up together, since the naive rearrangement OA/OB = OC/OD is a different and false statement. The angles AOD and COB are vertically opposite, so they are equal, and they are enclosed by exactly the sides in the proportion. SAS then gives triangle AOD similar to triangle COB, so ∠A = ∠C and ∠D = ∠B.
• Example 8 parts (i) and (ii), Fig. 6.33 (pp. 93–94, checked). Triangle ABC has median CM to side AB and triangle PQR has median RN to PQ, and ABC is given similar to PQR. Verified for part (i): the similarity gives AB/PQ = BC/QR = CA/RP and ∠A = ∠P; since M and N are midpoints, AB = 2 AM and PQ = 2 PN, so AM/PN = 2AM/2PN = AB/PQ = CA/RP. With ∠MAC equal to ∠NPR and those angles enclosed by the sides just paired, SAS gives triangle AMC similar to triangle PNR. Verified for part (ii): from that similarity CM/RN = CA/RP, and CA/RP already equals AB/PQ, so CM/RN = AB/PQ — the medians are in the same ratio as the sides. The idea worth stating: multiplying both terms of a ratio by 2 changes nothing, which is why a median can be swapped in for a side.
• Exercise 6.3 question 1 item (iv), Fig. 6.34 (p. 95, checked). Triangle MNL with NM = 2.5 and ML = 5 and the angle at M marked 70°; triangle QPR with QP = 5 and QR = 10 and the angle at Q marked 70°. Verified: 2.5/5 = 5/10 = ½, and in each triangle the 70° lies between exactly the two sides given. So SAS applies and MNL is similar to QPR.
• Exercise 6.3 question 1 item (v), Fig. 6.34 (p. 95, checked). Triangle ABC with AB = 2.5 and BC = 3 and the angle at A marked 80°; triangle DEF with DF = 5 and EF = 6 and the angle at F marked 80°. Verified: the ratios do agree, 2.5/5 = ½ and 3/6 = ½. But in DEF the 80° at F lies between the two given sides FD and FE, whereas in ABC the 80° at A lies between AB and AC, and AC is not among the given lengths — the two given sides AB and BC meet at B, not at A. So the enclosure condition fails and the criterion cannot be applied, even though every number in sight looks cooperative. This is the counter-case the whole topic should be built around, and it is the chapter's own, not an invented one.
• Exercise 6.3 questions 12 and 14, and Fig. 6.41 (p. 97, checked). Only question 12 carries the figure reference; question 14 is printed without one, though the drawing serves it equally well. Question 12: AB, BC and the median AD of one triangle are proportional to PQ, QR and the median PM of another; show the triangles are similar. Question 14: AB, AC and the median AD are proportional to PQ, PR and the median PM. The figure shows triangle ABC with D on BC and triangle PQR with M on QR, drawn side by side. The two questions differ in which pair of sides is named, and that difference is the whole exercise.
• A NOTE TO THE READER (p. 98, checked). After the summary the chapter adds a criterion for right triangles: if the hypotenuse and one other side of one are proportional to the hypotenuse and one other side of the second, the two are similar, and this may be called the RHS criterion. It points at Example 2 of Chapter 8 as the place it pays off — a deliberate cross-reference beyond this chapter's own folios. Worth flagging: this is a genuine addition to the list of criteria, sitting outside the numbered sections, and it is easy to miss because it comes after the chapter has apparently ended.

Figures to have open

• Item (v) of Fig. 6.34 (p. 95) redrawn with the two given sides highlighted in each triangle and the 80° arc drawn exactly where the book draws it. This is the topic's load-bearing figure and it only works if the angle's position is unmistakable.
• Item (iv) of the same figure, drawn the same way, as the passing case.
• Fig. 6.28 (p. 90) with the dashed segment PQ, shown step by step so triangle DPQ can be lifted out and laid over ABC — the same reusable asset as the other two criteria.
• Fig. 6.31 (p. 92): two segments crossing at O, lettered so that the two triangles AOD and COB can be lit separately. The crossing must look like a crossing.
• Fig. 6.33 (p. 93) with both medians drawn and the midpoints marked with equal tick pairs.
• A hinge movement for section 4: two rods of fixed length joined at a pivot, with the third side drawn as the pivot opens. Standard schematic, and the one visual that explains why enclosure matters.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 6 "Triangles", §6.4, pp. 89–91: Activity 6, Theorem 6.5 and its proof. Figures 6.27 and 6.28.
• Example 6, p. 92, with Fig. 6.31; Example 8, pp. 93–94, with Fig. 6.33.
• Exercise 6.3, pp. 94–97: question 1 items (iv) and (v) with Fig. 6.34, and questions 12, 14 and 16 with Fig. 6.41.
• Item 9 of the chapter's summary, p. 98, and the A NOTE TO THE READER box on the same page, which is the source for the RHS criterion and which points at Chapter 8 — outside pp. 73–98 and cited here as the chapter's own cross-reference.