Euclidea Wiki

Instruction: Erect a perpendicular from the point on the line.

Goal: 1L 3E

Available tools:

  • Move
  • Point
  • Line
  • Circle
  • Perpendicular Bisector
  • Angle Bisector
  • Intersect

Pack: Beta

Previous level: 2.6

Next level: 2.8

1L solution[]

Let O be the point, and A and B two arbitrary points on the line so that O is between them.

  1. Construct the angle bisector of ∠AOB

3E solution[]


Let O be the point.

  1. Construct a circle with an arbitrary center A not on the line and radius OA, intersecting the line at B
  2. Construct line AB, intersecting circle A at C
  3. Construct line OC


Solution 1L is simple as a line is an angle of 180°, so its angle bisector will divide it into two angles of 90° and will therefore be perpendicular to it. Just have the vertex of the 180° angle be point O.

Solution 3E makes use of Thales's theorem. You are constructing the inscribed triangle OBC with its side BC a diameter of the circle, which makes OBC a right triangle with ∠BOC=90°.