# Perpendicular to a line from an external point

This page shows how to construct a perpendicular to a line through an external point, using only a compass and straightedge or ruler. It works by creating a line segment on the given line, then bisecting it. The bisector will be a right angles to the given line. (See proof below).

## Printable step-by-step instructions

The above animation is available as a printable step-by-step instruction sheet, which can be used for making handouts or when a computer is not available.

## Proof

The image below is the final drawing above with the red lines added. Argument Reason
1 Segment RP is congruent to RQ They were both drawn with the same compass width
2 Segment SQ is congruent SP They were both drawn with the same compass width
3 Triangle RQS is congruent to triangle RPS Three sides congruent (sss), RS is common to both.
4 Angle JRQ is congruent to JRP CPCTC. Corresponding parts of congruent triangles are congruent.
5 Triangle RJQ is congruent to triangle RJP Two sides and included angle congruent (SAS), RJ is common to both.
6 Angle RJP and RJQ are congruent CPCTC. Corresponding parts of congruent triangles are congruent.
7 Angle RJP and RJQ are 90° They are congruent and supplementary (add to 180°).

- Q.E.D

## Try it yourself

Click here for a printable construction worksheet containing two 'perpendiculars through a point' problems to try. When you get to the page, use the browser print command to print as many as you wish. The printed output is not copyright.