‹ Class 7 · Ch 14
Constructions and Tilings · Principle 2 of 14

Two points equally far from X and Y

AX = AY = BX = BY makes AB the perpendicular bisector.

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NCERT: 6.1 Geometric Constructions

Think

Four equal distances

Mark two points A and B, on opposite sides of XY, so that AX = AY = BX = BY. This is what makes the eye symmetrical.

Join A and B. Where will the line AB meet XY?

What this lesson covers

The idea

If A and B, on opposite sides of XY, satisfy AX = AY = BX = BY, then the line AB is the perpendicular bisector of XY; this follows from congruent triangles (SSS, then SAS).

Four equal distances

Mark two points A and B, on opposite sides of XY, so that AX = AY = BX = BY. This is what makes the eye symmetrical.

Join A and B. Where will the line AB meet XY?

  • In the middle of XY, at a right angle
  • In the middle of XY, but at a slant
  • Somewhere that depends on the radius

Change the radius, change XY

Arcs of the same radius from X and from Y meet at A and B. The figure shows what AB does to XY. Change the radius and the length of XY, then fold along AB.

Why it always works

It worked for every radius. Congruent triangles show why.

The angles ∠AOX and ∠AOY are equal and together form a straight angle, 180°. So each is 90°.

If A and B, on opposite sides of XY, satisfy AX = AY = BX = BY, then the line AB is the perpendicular bisector of XY. It follows from congruent triangles (SSS, then SAS).

  • Triangles | Why they match
  • ∆ABX and ∆ABY | AX = AY, BX = BY, and AB is common: SSS. So ∠XAB = ∠YAB.
  • ∆AOX and ∆AOY | AX = AY, ∠XAO = ∠YAO, and AO is common: SAS. So OX = OY and ∠AOX = ∠AOY.

Notes

If A and B, on opposite sides of XY, satisfy AX = AY = BX = BY, then the line AB is the perpendicular bisector of XY: congruent triangles (SSS, then SAS).

Check yourself

A and B are on opposite sides of XY, and AX = AY = BX = BY = 5 cm. XY is 6 cm. AB crosses XY at O. How many cm is XO?

Answer: 3

AB is the perpendicular bisector of XY, so O is the midpoint: XO = 6 ÷ 2 = 3 cm.

To show that ∆ABX ≅ ∆ABY we know AX = AY, BX = BY and that AB is a common side. Which congruence condition is that?

∆AOX ≅ ∆AOY, so ∠AOX = ∠AOY. They lie on the straight line XY, so together they make 180°. How many degrees is ∠AOX?

Answer: 90

Two equal angles that together make 180° are 180° ÷ 2 = 90° each. So AB is perpendicular to XY.

Anu says: “AB is the perpendicular bisector only if the arcs have a radius exactly equal to XY.” Is she right?

  • SSS (three pairs of equal sides) — correct. Yes! The three pairs of equal sides are AX and AY, BX and BY, and AB with itself.
  • SAS (two sides and the angle between them). We have no angle yet. In fact, we use this triangle pair to get the equal angles.
  • ASA (two angles and the side between them). We know no angles at this stage, only sides.
  • Yes. Other radii give a slanted line.. Every radius you tried gave a right angle at the midpoint. What matters is AX = AY = BX = BY.
  • Yes. Other radii move the midpoint.. The midpoint of XY is fixed by X and Y. The radius does not move it.
  • No. Any radius that lets the arcs meet works, since AX = AY = BX = BY still holds. — correct. Yes! In the toy, every radius gave the same result: AB through the midpoint at 90°.
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