FG2, HG2 - Ruler and compass constructions

Perpendicular bisector

The perpendicular bisector of a segment crosses it at its midpoint at $90^\circ$. Every point on this line is equally distant from the segment's two endpoints.

Worked example 1: bisecting a segment

Question. Construct the perpendicular bisector of a segment $AB$ of length $6\,\mathrm{cm}$.

Plan. Use equal compass radii greater than $3\,\mathrm{cm}$, then join the two intersections.

1. Draw $AB=6\,\mathrm{cm}$ with a ruler.
2. Set the compass to $4\,\mathrm{cm}$. Draw arcs above and below $AB$, centre $A$.
3. Keep the $4\,\mathrm{cm}$ radius and repeat from $B$. Label the two arc intersections $X$ and $Y$.
4. Use the ruler's straight edge to draw $XY$, leaving the arcs.

Answer. $XY$ is the required perpendicular bisector.

Check. After constructing, a ruler should show two $3\,\mathrm{cm}$ halves, and a set square or protractor can check the right angle. These checks verify the drawing; they do not replace the compass construction.

Worked example 2: perpendicular at a point

Question. Construct a perpendicular to line $l$ through a point $P$ on $l$.

Plan. Mark points equally far from $P$, then construct their perpendicular bisector.

1. With centre $P$, draw an arc crossing $l$ at $U$ and $V$. The compass makes $PU=PV$.
2. From $U$ and $V$, use equal radii greater than $UP$ to draw intersecting arcs above the line at $Q$.
3. Draw $PQ$. Since $P$ is the midpoint of $UV$, $PQ$ is perpendicular to $l$.

Answer. The straight line $PQ$ is the required perpendicular.

Check. It passes through the original point $P$, not merely a nearby point, and both angles next to it on $l$ should be $90^\circ$.

Worked example 3: perpendicular from a point

Question. Construct a perpendicular from a point $P$ above line $l$ to $l$.

Plan. Use an arc centred at $P$ to create two points on $l$ equally distant from it.

1. Open the compass far enough that an arc centred at $P$ cuts $l$ at two distinct points $U$ and $V$. Extend the line first if needed.
2. From $U$ and $V$, use one equal radius large enough for the arcs to meet below $l$ at $Q$.
3. Draw $PQ$, crossing $l$ at $H$.

Both $P$ and $Q$ are equidistant from $U$ and $V$, so $PQ$ is their perpendicular bisector.

Answer. $PH$ is the perpendicular from $P$ to $l$.

Check. The constructed line must go through $P$, meet $l$, and make a right angle there. The first compass arc need not have the same radius as the later pair; only the two arcs from $U$ and $V$ must share a radius.

Watch out

Changing the radius between centres $A$ and $B$ breaks $XA=XB$. Use one compass setting for the matched arcs. A line drawn through the estimated midpoint is not a ruler-and-compass construction.

Angle bisector

An angle bisector is a ray from the vertex that divides the angle into two equal parts. For example, it divides $68^\circ$ into $34^\circ$ and $34^\circ$.