The perpendicular bisector, and the equidistance property that justifies it

What to assume they know

Drawing an arc of a chosen radius from a chosen centre with a compass
Triangle congruence, and reading off equal parts once two triangles match (Part II, printed Chapter 1, "Geometric Twins")
The SAS condition, and knowing which angle is the included one (SAS, and why the angle has to be the included one)
Angles on one side of a straight line add to 180° (The four angles at a crossing: vertically opposite and linear pairs; Part I, printed Chapter 5 "Parallel and Intersecting Lines", §5.1)
The Class 6 "Eyes" construction, in which two arc centres were placed by eye

What they should be able to do

State what bisection means and what the extra word perpendicular adds
Explain why the four arcs of the construction produce points that are the same distance from both ends of the segment
Reconstruct the two-step congruence argument and say which condition each step uses
Explain why two equal angles that together fill a straight angle must each be 90°
Justify the converse: a point equally far from both endpoints has to sit on their perpendicular bisector
Carry out the construction with a compass and an unmarked ruler
Decide, with a reason, whether the arcs above and below need the same radius
Say why this construction locates a midpoint more reliably than measuring does

Where it usually goes wrong

"The arcs above and below must be drawn with the same radius." This is the first thing the chapter's own Figure it Out attacks (p.140, question 1). They need not. Any point above that is equally far from both endpoints will do, and any point below likewise; two such points fix the line.
"You need all four arcs — that is what makes it work." Four arcs give two points, and two points are what a line needs. The chapter makes the same observation do real work on p.141, where one of the two points is already known and the second pair of arcs is dropped.
"Both crossings have to be on opposite sides of XY." Question 2 on p.140 asks exactly this and invites a construction rather than a rule. Nothing in the argument mentions sides; it only mentions distances.
"It is perpendicular because it looks perpendicular." The whole of pp.137–138 exists because looking is not enough. The right angle is deduced from two equal angles that fill a straight angle, not measured off the drawing.
"Bisecting is just finding the middle by measuring." The chapter says the compass method is the more accurate of the two (p.141). Measuring introduces a reading error twice over; the compass never reads anything.
"Bisection is a thing you do to line segments." The definition on p.137 is wider than that, and the chapter cashes it in two pages later on angles. Keep the definition general when you state it.

Questions to check understanding

Construct the perpendicular bisector of a given segment and state the two tools used
Given a point marked equally far from both endpoints of a segment, say what can be concluded about where it lies, and why
Justify that the drawn line really is perpendicular, naming the congruence conditions used (the shape of the question on p.137)
Decide whether the arcs above and below may have different radii, and defend the answer by construction (Figure it Out question 1, Part II, p.140)
Decide whether both pairs of arcs may be drawn on the same side of the segment (Figure it Out question 2, Part II, p.140)
Reproduce a four-petal or four-leaf design using only a compass and an unmarked ruler (Figure it Out question 4, Part II, p.140)
The bisection bullet and the equal-distance bullet of the SUMMARY (Part II, p.163) are the two statements the chapter expects back

Examples worth working on the board

The four equal lengths (Part II, §6.1, p.136). X and Y are the two corners of the eye; A is the centre of the upper arc and B the centre of the lower one. For the two arcs to match, AX must equal BX, and since a circle's radius is constant, AX = AY and BX = BY already. The chapter chains these into one string of four equal lengths.
Fig. 6.1 (Part II, §6.1, p.137). Checked against the printed page. Two small crosses, one above the segment XY and one below, where the arc pairs meet; the upper crossing is lettered A and the lower one B. In the figure printed just after it, A and B are joined and the join meets XY at a point lettered O, so the drawing becomes a kite-shaped quadrilateral XAYB with both diagonals drawn.
The straight-angle step (Part II, §6.1, p.137). Inputs: ∠AOX and ∠AOY are equal, and together they make a straight angle, so their sum is 180°.
The second pair of centres (Part II, §6.1, p.139). Checked against the printed page. Points C and D are marked with the same relation to X and Y that A and B had, and the printed figure stacks them on one vertical line in the order C, A, X–Y, B, D — which is the visual claim that all four lie on a single line.
The three construction steps (Part II, §6.1, pp.139–140). Checked against the printed page. Step 1 (p.139) shows only the segment XY with a crossing above it lettered A. Step 2 (p.140) shows the finished figure: crossings above and below, lettered A and B, joined by a line that cuts XY. Step 3 names AB as the answer. The chapter binds itself to these two tools for the rest of the chapter except where a length in standard units is called for — that exception is printed alongside the restriction at the foot of p.140, and it is what the centimetre hexagons of pp.152–153 lean on.
The four-petal design (Part II, §6.1, "Figure it Out", question 4, p.140). Checked against the printed page: four congruent leaf shapes meeting at one point, one pointing up, one down, one left, one right, each bounded by two arcs. It is a ruler-and-compass exercise, and it is the right closing image.
The accuracy claim (Part II, §6.1, p.141). The chapter states plainly that locating a midpoint this way beats reading the segment off a marked scale.

Figures to have open

The eye outline with X and Y at the corners and the two arcs meeting there. Standard schematic; it should be able to be shown moving so the arc centres can be moved.
The four-arc construction, able to be shown moving step by step, with the crossings appearing as the arcs sweep. Standard schematic.
The kite XAYB with both diagonals, so that ΔABX / ΔABY and ΔAOX / ΔAOY can each be lifted out and set side by side. This is the topic's key image.
A vertical line carrying several equidistant centres (C, A, B, D) with a family of differently shaped eyes drawn from them, for section 10.
The four-petal design of p.140 question 4, redrawn.
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 6 "Constructions and Tilings", §6.1 "Geometric Constructions" — the unnumbered subheading "Eyes" and the four equal lengths (p.136), Fig. 6.1 and the naming of bisection and the perpendicular bisector (p.137), the two congruences (pp.137–138), the points C and D and the equal-distance statement (p.139)
Same part, same chapter, §6.1, "Construction of Perpendicular Bisector", pp.139–140, and the two-tools remark at the foot of p.140
Same part, same chapter, §6.1, "Figure it Out", questions 1–4, p.140
Same part, same chapter, §6.1, the accuracy remark, p.141
Same part, same chapter, SUMMARY, p.163, the bisection bullet and the equal-distance bullet
Forward pointers: Constructing a 90° angle at a chosen point on a line (the same construction turned round to make a right angle) and Bisecting an angle, and halving 90° to get 45° (the same idea applied to an angle instead of a segment)
Backward pointer: SAS, and why the angle has to be the included one for the SAS condition this argument leans on