RHS: the right-angled special case

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What to assume they know

SAS, and why the angle has to be the included one — SAS, SSA, and the two-crossings construction
SSS: three sidelengths fix a triangle completely — SSS, and the flip argument about a base line
Drawing a line perpendicular to a given line through a given point
A right angle measures 90°, and a triangle can contain at most one
The idea that the largest side of a triangle faces its largest angle (assumed, not proved here)

What they should be able to do

State the RHS condition, naming which three parts have to match
Identify the hypotenuse in a right-angled triangle
Carry out the four-step construction from a base, a perpendicular and an arc
Explain why the arc necessarily meets the perpendicular on both sides of the base
Argue that the two resulting triangles are congruent, using the base as a fold line
Say precisely how this case differs from the SSA counterexample
Recite the chapter's list of five sufficient conditions, and say what is not on it
Apply RHS to a labelled pair of right-angled triangles

Where it usually goes wrong

"RHS proves that SSA works after all." It does not. SSA still fails, and the chapter keeps that failure on the books. What RHS says is that this particular SSA-shaped situation is safe, for a reason you can see in the picture.
"Any two sides of a right-angled triangle will do." One of the two has to be the hypotenuse. Give a right angle and the two shorter sides instead and you are in the SAS case, which also works — but for a different reason, and the chapter is careful to name only the hypotenuse version RHS.
"The hypotenuse is the bottom side" or "the slanted one." It is whichever side faces the right angle, wherever the triangle happens to be sitting on the page. The p.16 sketch has it slanting and the p.20 exercise has it named only by its letters — use both.
"The arc might miss the line below the base." It cannot. The base is perpendicular to l at Q, so Q is the closest point of l to R; whatever distance the arc reaches beyond that, it reaches equally far up and down. Say this in words, without a formula.
"A second solution always means the givens were not enough." This is the belief the whole chapter is training against, and it has now been wrong twice and right once. What matters is whether the second solution is a flip of the first.
"There are five conditions, so there are five things to memorise." AAS reduces to ASA by the angle sum, and RHS is a repaired SSA. Two of the five are consequences, not axioms.

Questions to check understanding

Identify the hypotenuse in a labelled right-angled triangle
Given a right angle, a hypotenuse and one other side, construct the triangle
Decide whether a stated set of equalities is an RHS set, an SAS set, or neither
Explain why the two triangles produced above and below the base are congruent
Given two right-angled triangles with marked equal parts, express the congruence (the shape of Figure it Out question 2, case (c), Part II, p.20)
Explain why RHS does not contradict the general failure of SSA
The hypotenuse bullet and the RHS bullet of the SUMMARY (Part II, p.22) are the chapter's own compressed statements
RHS is the condition used in Part II, §1.3 to prove the isosceles-angle result, so it is examined again indirectly there

Examples worth working on the board

The RHS case (Part II, §1.2, p.16). Inputs: BC and YZ are both 4 cm; ∠B and ∠Y are both 90°; AC and XZ are both 5 cm. Note which side is which — AC faces the right angle at B, so it is the hypotenuse, while BC is one of the two shorter sides. That distinction is the condition.
The rough diagram (Part II, §1.2, p.16). Checked against the printed page. A right triangle with P at the top, Q at the lower left carrying a small square right-angle mark, and R at the lower right. QR is tagged 4 cm along the bottom and PR is tagged 5 cm on the slope. Note the relabelling: P, Q, R here play the parts of A, B, C in the problem statement.
The four steps (Part II, §1.2, p.16). Checked against the printed page. Step 1 draws QR = 4 cm. Step 2 raises l at right angles to QR, standing at Q; the printed figure shows it running well above Q with an arrowhead. Step 3 swings an arc of radius 5 cm centred on R until it crosses l. Step 4 marks the crossing as P and joins PR.
The third side. With the right angle at Q, the base 4 cm and the hypotenuse 5 cm, the remaining side PQ comes out 3 cm. That number is added here, computed from the book's three inputs; the chapter prints neither it nor any relation among the three sides. It is here only so the figure can draw the figure to scale, and it must not be spoken as something the chapter supplies.
The downward extension (Part II, §1.2, p.17). The chapter's own follow-up question: continue l below QR and ask whether the arc from R meets it there too, and if so whether that gives a different triangle. Inputs: the same construction, one line extended. The answer the chapter gives is that the second triangle is congruent to the first — and it then leaves the why to the reader. That "why" is the explanation's job, and the answer is the flip about QR.
The two equal parts, as the chapter lists them (Part II, §1.2, p.17). It itemises what matches: the right angle, then two further sides, one of them the one facing that right angle. The naming of the hypotenuse arrives inside the second item.
The list of five (Part II, §1.2, p.17). Checked against the printed page. Under a bold unnumbered heading the chapter sets out SSS, SAS, ASA, AAS and RHS as five lettered items in a two-column block. SSA is absent from the list, deliberately. Show the list and show the gap.
Figure it Out, question 2, case (c) (Part II, §1.3, p.20). Checked against the printed page. Inputs: AB = DF, ∠B = ∠D = 90°, AC = FE. Both right angles are at B and at D; AC faces ∠B and FE faces ∠D, so both are hypotenuses. This is the chapter's own RHS exercise item. The congruence that follows is left to the reader.

Figures to have open

A right-angled triangle whose right angle can be marked with the small square, with the hypotenuse distinguishable from the other two sides. Standard schematic.
The base–perpendicular–arc construction, able to be shown moving in both directions so the perpendicular can extend below the base and the arc can cross it twice. This is the topic's central image.
A fold movement about the base, reusing the one built for SSS: three sidelengths fix a triangle completely so the parallel with SSS is visible.
A side-by-side of the p.12 SSA construction and this one, for section 7. Not a printed figure; assemble it.
A five-item list of the sufficient conditions with SSA shown excluded.
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.16–17 — the bridge from AAS and the note that SSA has special cases (p.16), the unnumbered subheading "Measuring Two Sides in a Right Triangle" with the rough diagram and the four steps (p.16), the downward-extension question, the two itemised equal parts, the naming of the hypotenuse and of RHS (p.17), and the bold unnumbered heading over the list of five sufficient conditions (p.17)
Same part, same chapter, §1.3, "Figure it Out", question 2, case (c), p.20
Same part, same chapter, SUMMARY, p.22, the hypotenuse and RHS bullets
Backward pointer: SAS, and why the angle has to be the included one, whose counterexample this topic reads against
Forward pointer: Angles opposite equal sides are equal, where RHS does the work in Part II, §1.3