Bisecting an angle, and halving 90° to get 45°

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What to assume they know

The perpendicular bisector, and the equidistance property that justifies it — the perpendicular bisector and the equal-distance property
Constructing a 90° angle at a chosen point on a line — building a 90° angle at a chosen point
The SSS condition for triangle congruence (SSS: three sidelengths fix a triangle completely)
A complete turn about a point measures 360°
Reading equal parts off a pair of congruent triangles

What they should be able to do

Work out, from an eight-armed design, that its supporting lines must sit 45° apart
Explain why 45° has to be reached by halving rather than by drawing
Describe the pair of triangles whose congruence would bisect a given angle
Explain how marking OA = OB and cutting equal arcs makes those triangles congruent by SSS
Carry out the three-step bisection on any drawn angle
Construct 45° from 90°, and use it to build the eight-armed figure
Investigate what happens when the arcs are cut on the far side of the vertex
Say which angles repeated bisection can and cannot reach

Where it usually goes wrong

"Bisecting an angle is a different skill from bisecting a segment." Both are the same move: build equal lengths so that two triangles must be congruent, then read the equality you wanted off the congruence. Say this out loud; it is the chapter's design.
"The arcs from A and B must have the radius used to mark A and B." They need not. The chapter's step 2 says only that the radius must be long enough and the same for both arcs.
"C has to lie between the arms." Question 3 exists to unsettle exactly that. The two possible crossing points and the vertex are collinear, so the line is unchanged; what changes is which angle the ray on that side sits inside.
"A protractor would be quicker, so this is just a school exercise." The chapter has restricted itself to two tools since p.140, and the point is that the result is exact rather than read off a scale.
"Repeated bisection will reach any angle you like." It reaches halves of halves. Question 4 is placed there to make a student notice the limit, not to be answered by "yes".
"Eight arms must mean eight separate constructions." Four lines through one point give eight rays. Building 90° once and bisecting twice gets the whole skeleton.

Questions to check understanding

Bisect a given angle with a compass and an unmarked ruler, and state the congruence condition that justifies it
Construct a 45° angle, showing the 90° stage
Construct several angles of different sizes and bisect each (Figure it Out question 1, Part II, p.144)
Reproduce the eight-petalled figure (Figure it Out question 2, Part II, p.144)
Decide whether the bisector is unchanged when the arcs are cut on the other side, and defend the answer (Figure it Out question 3, Part II, p.144)
List angles obtainable by repeated bisection, and test a given value against the list (Figure it Out question 4, Part II, p.144)
Construct a design of petals of the largest possible size inside a given square (Figure it Out question 6, Part II, p.145)
The bisected-and-copied bullet of the SUMMARY (Part II, p.163) is the statement the chapter expects back

Examples worth working on the board

Fig. 6.5 (Part II, §6.1, p.143). Checked against the printed page. Eight identical narrow leaf shapes radiating from one centre, evenly spaced, each bounded by two arcs, so the outline reads as a rosette or an eight-petalled flower. Directly beneath it the chapter prints its skeleton: four straight lines crossing at one point, with dotted lines between them.
The 45° calculation (Part II, §6.1, p.143). Inputs: a full turn is 360°, and the design divides it into eight equal parts.
The bisection target (Part II, §6.1, p.143). Checked against the printed page. The chapter prints ∠XOY with O at the vertex, OY to the right and OX rising to the upper right, then reprints it with A marked on OY, B marked on OX, and C further out between the two arms, so that ΔOBC and ΔOAC sit side by side sharing OC.
The three conditions (Part II, §6.1, p.144). Inputs: OA = OB; BC = AC; OC is common to both triangles. Those three are what SSS needs.
The printed steps (Part II, §6.1, p.144). Checked against the printed page. Step 1 marks A on OY and B on OX with two short compass ticks of one radius, so that OA = OB; no single arc is swept across both arms, and the printed instruction asks only for the two points at equal distance. Step 2 adds two arcs of one radius struck from A and from B, meeting at C between the arms. Step 3 draws OC. The angle mark and the equal-length ticks are drawn on the figures, not set in the text.
The far-side variant (Part II, §6.1, question 3, p.144). Checked against the printed page. Same angle, same A and B, but the two arcs are struck on the other side of O so that C lands out to the left of O, past the vertex.
The 65.5° question (Part II, §6.1, question 4, p.144). The chapter asks which further angles bisection reaches, and whether 65.5° is one of them. It prints no answer. See Notes before letting the explanation say anything about it.
Two designs (Part II, §6.1, p.144 question 2 and p.145 question 6). The first is Fig. 6.5 itself. The second, checked against the printed page, is a square with four petals inside it: each petal is bounded by two arcs, the four meet at the centre of the square, and the chapter adds the harder question of making them as large as the square allows.

Figures to have open

The eight-petalled rosette and its four-line skeleton, able to be shown moving so the skeleton can fade in under the finished figure. Redraw rather than reproduce.
An angle with movable arms, carrying A, B and C, so the bisection can run on any opening. This is the topic's key image.
The pair ΔOBC / ΔOAC lifted out of that figure and set side by side, built fresh — the chapter draws them only nested inside the angle.
The far-side variant with C beyond the vertex, and a single straight line drawn through O, C and the between-the-arms crossing point, to show they are collinear. Assembled by the explanation; the chapter prints only the figure, not the line.
The square-with-four-petals design of p.145.
No photograph, table or dataset from the textbook is needed.

Where this sits in the book

NCERT Ganita Prakash, Class 7, Part II, printed Chapter 6 "Constructions and Tilings", §6.1 "Geometric Constructions", the unnumbered subheading "Angle Bisection for a Design", p.143, with Fig. 6.5 and its skeleton
Same part, same chapter, §6.1, "Steps for Angle Bisection", p.144
Same part, same chapter, §6.1, "Figure it Out", questions 1–5, p.144, and question 6, p.145
Same part, same chapter, SUMMARY, p.163, the bisecting-and-copying bullet
Backward pointers: The perpendicular bisector, and the equidistance property that justifies it, Constructing a 90° angle at a chosen point on a line
Forward pointers: Copying an angle, and why triangle congruence proves it works (copying an angle, the other half of the SUMMARY bullet) and Constructing 60° from an equilateral triangle, and the arches built on it (30° and 15°, this method applied to 60°)