__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.

- Construct the angle bisector of ∠AOB

## 3E solution[]

Let O be the point.

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

## Explanation[]

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°.