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The Perpendicular Bisector

Part of the Transformations & Constructions section of Coddy's Geometry journey. Lesson 41 of 60.

The perpendicular bisector of a segment crosses it at its midpoint, at right angles. Every point on it is the same distance from both ends, and every point that is equally far from both ends lies on it.

$Circles about X and Y cross at U and V; the line UV is the perpendicular bisector of XY$

To build it for XY: draw the circle about X through Y and the circle about Y through X. They cross at U and V, and the line UV is the perpendicular bisector. Why? U is XY from X and XY from Y, so it is equally far from both ends, and so is V. Better still, XUYV has four sides all equal to XY: it is a rhombus, and the diagonals of a rhombus cut each other in half at right angles.

Challenge

Construct the perpendicular bisector of AB, then press Check.

Try it yourself

$Construct it with a compass and straightedge$

Construct it with a compass and straightedge

All lessons in Transformations & Constructions

1Translations

Sliding a Shape Draw the Image The Translation Rule Recap: Translations

4Combining Moves

One Move, Then Another Order Matters Two Mirrors Name the Move Recap: Combining Moves

7Drawing on Squared Paper

Lengths by Pythagoras Perpendicular Steps Squares by Area Right Triangles by Area Midpoints on the Grid Recap: Squared Paper

10Perpendiculars and Parallels

Perpendicular From a Point Perpendicular at a Point Parallel Through a Point Tools You Have Earned Recap: Parallels

3Rotations

Turning About a Point The Half Turn Quarter-Turn Rules A Center Off the Origin Recap: Rotations

6Symmetry

Lines of Symmetry Rotational Symmetry Regular Polygons Symmetry Is a Move Recap: Symmetry

9Bisectors

The Perpendicular BisectorThe Midpoint Bisecting an Angle Half of 60°Recap: Bisectors

12Final Challenges

Final: Footprints Final: Two Moves, One Rule Final: Mirror on the Grid Final: The Fair Well Final: The Tangent Road