Angles opposite equal sides are equal

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What to assume they know

RHS: the right-angled special case — the RHS condition and the naming of the hypotenuse
Naming congruent figures so the correspondence is unambiguous — reading corresponding parts off a congruence statement
Isosceles means two sides of equal length
Dropping a perpendicular from a vertex to the opposite side
The three angles of a triangle add to 180° — Why the three angles of any triangle add to 180° (Part I, printed Chapter 7 "A Tale of Three Intersecting Lines", pp.165–166)
Every radius of a circle has the same length

What they should be able to do

State the isosceles result: equal sides force the angles facing them to be equal
Construct the altitude from the apex of an isosceles triangle to its base
List the three equalities the altitude creates, and say where each comes from
Identify RHS as the condition that applies, and say which side is the hypotenuse
Deduce the equality of the base angles from the correspondence
Explain why the argument holds for every isosceles triangle, not just for the printed one
Compute the two base angles when the apex angle is given
Apply the result to a triangle formed by two radii of a circle

Where it usually goes wrong

"You can only use congruence when there are two triangles in the picture." This section exists to break that. There is one triangle on p.17 and two on p.18, and the reader put the second one there. Drawing an extra line is a legitimate move, and it is the single most transferable idea in the chapter.
"Equal angles because the picture looks symmetrical." The picture does look symmetrical, which is exactly why the argument matters. The book could have said "fold it and see"; instead it proves it, and the proof is what carries over to cases where the symmetry is not visible — the circle on p.21, for instance.
"∠B and ∠C are equal because they are both at the bottom." They are equal because AB and AC are equal. Turn the triangle on its side in the figure and keep the tick marks; the conclusion has to travel with the marks, not with the page.
"The altitude was chosen because it looks like the axis of symmetry." It was chosen because it manufactures two right angles, which is what RHS needs. Say what the construction is for.
"The result depends on the apex being 80°." Section 8 has to make the opposite point out loud. Nothing in the argument used the number; the 80° is there only so that the follow-up question has an answer.
"Angles opposite equal sides are equal" reversed — "equal angles mean equal sides." That is a different statement. The chapter does not prove it.
"Congruence proves the two halves are the same triangle." They are two different triangles that happen to match part for part. The correspondence A↔A, D↔D, B↔C is what licenses reading ∠B against ∠C, and it should be shown.

Questions to check understanding

Given an isosceles triangle and its apex angle, find the other two angles
Given an isosceles triangle and one base angle, find the apex angle
Prove that the angles facing two equal sides are equal, naming the construction and the condition used
Given a triangle formed by two radii of a circle, find its other two angles (Figure it Out question 5, Part II, p.21)
Show that a drawn segment cuts an angle into two equal pieces by proving the triangles either side of it congruent (Figure it Out question 3, Part II, p.9)
Find missing angles in a figure built from several marked-equal segments (Figure it Out question 6, Part II, p.21)
The angles-opposite-equal-sides bullet of the SUMMARY (Part II, p.22)

Examples worth working on the board

The isosceles triangle (Part II, §1.3, p.17). Checked against the printed page. ΔABC is drawn with A at the apex, B at the lower left and C at the lower right. A single tick sits on AB and a matching tick on AC, and an arc at A is labelled 80°. Inputs: AB = AC and the apex angle is 80°. Note a typographic detail — in the body text on p.17 the degree sign after the 80 is missing, while the figure carries it. Checked against p.17. The intended value is 80°.
The altitude (Part II, §1.3, pp.17–18). Checked against the printed page. The reader is told to construct the altitude from A to BC; the figure on p.18 shows the same triangle with D marked on BC, AD drawn vertically, and a small square right-angle mark at D. AD is the added line — everything that follows depends on it.
The three equalities (Part II, §1.3, p.18). Inputs, in the chapter's own order: AB = AC because it was given; the two angles at D are both 90° because that is what the construction made them; AD belongs to both halves. Note which sides are which: AB and AC face the right angles at D, so they are the hypotenuses of the two halves, and AD is the shared other side.
The conclusion (Part II, §1.3, p.18). ΔADB ≅ ΔADC, and ∠B and ∠C are corresponding parts, so they are equal. The general statement follows: in any triangle, the angles facing equal sides are equal.
Finding the base angles (Part II, §1.3, p.18). Inputs: apex 80°, three angles totalling 180°, and the two base angles now known to be equal. The chapter asks for ∠B and ∠C and does not print the answer. Do not hand it the answer, and do not present 50° as a printed value.
Figure it Out, question 5 (Part II, §1.3, p.21). Checked against the printed page. A circle with A marked at its centre and B and C sitting on the circle, with AB, AC and the segment BC drawn. The angle at A is labelled 120°. Inputs: A is the centre, so AB and AC are radii and therefore equal; the apex angle is 120°; the three angles total 180°. This is the same argument with a different apex value.
Figure it Out, question 3 (Part II, §1.2, p.9). Checked against the printed page. The four-cornered figure with A at the top, B at the left, C at the bottom and D at the right, AC dashed across it, AB = AD and CB = CD marked. Its second half — asking whether AC halves each of ∠BAD and of ∠BCD — runs on this topic's reasoning: prove the congruence, then read off the parts.
The chapter's compressed statement (Part II, SUMMARY, p.22). The second-to-last bullet is this result in one line. Use it as the recap.

Figures to have open

An isosceles triangle with the two equal sides tick-marked and a labellable apex angle. Standard schematic; it must be re-drawable at several apex values for section 8.
A split-and-hinge movement of that triangle about its altitude, with the right-angle mark appearing at the foot. This is the topic's central image.
A circle with a marked centre, two radii and the segment joining their ends, for section 10. Standard schematic.
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", §1.3 "Angles of Isosceles and Equilateral Triangles", pp.17–18 — the opening claim that congruence is a tool, the isosceles triangle with its 80° apex and the instruction to construct the altitude (p.17), and the three equalities, the RHS verdict, the congruence of the two halves and the general statement (p.18)
Same part, same chapter, §1.3, "Figure it Out", questions 5 and 6, p.21
Same part, same chapter, §1.2, "Figure it Out", question 3, p.9, for the same reasoning applied to a four-cornered figure
Same part, same chapter, SUMMARY, p.22
Backward pointer: RHS: the right-angled special case for RHS
Forward pointer: Why every equilateral triangle has three 60° angles, which applies this result twice over