Standard Ruler and Compass Constructions

Perpendicular Bisectors

Understanding how to find the exact middle of a space explains why mobile phone masts are placed to cover overlapping territories equally. A perpendicular bisector is a line that divides a segment into two equal halves and meets it at exactly a $90^{\circ}$ angle. This creates a locus of points that are entirely equidistant (an equal distance) from the two endpoints of the line segment. First, place the compass point on endpoint $A$ of the line segment. Next, open the compass to a radius greater than half the length of the line. Then, draw an arc above and below the line. Crucially, keep the same setting and move the compass point to endpoint $B$. Draw arcs above and below to intersect the first set, labeling the intersections $P$ and $Q$. Finally, draw a solid line through $P$ and $Q$ using a straight edge. If applying this to coordinate geometry, you can find the bisector algebraically using the midpoint $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$ and the perpendicular gradient $-\frac{1}{m}$. Substitute these into the equation $y - y_1 = m(x - x_1)$.

Angle Bisectors

You can split any angle perfectly in half without ever measuring the angle itself! An angle bisector is a line that divides an angle into two equal, congruent parts. To bisect an angle is to find the locus of points equidistant from the two intersecting lines. First, place the compass on the vertex $B$ (the corner of the angle). Then, draw an arc crossing both lines forming the angle and label the intersection points $P$ and $Q$. Next, place the compass on $P$ and draw an arc in the center of the angle. Keeping the exact same width, place the compass on $Q$ and draw an arc intersecting the previous one. Label this $X$. Finally, draw a straight line from vertex $B$ through point $X$.

Perpendicular from a Point ON a Line

How do you draw a perfectly straight line upwards from a flat horizon using only circles? A perpendicular line drawn from a specific point on a base line uses the same logic as a perpendicular bisector by creating two temporary equidistant points. First, place the compass on the given point $P$ on the line. Then, draw two arcs of the exact same radius crossing the line on both sides of $P$. Label these $A$ and $B$. Next, open the compass to a wider setting (must be $> 0.5 \times AB$). Place the compass on $A$ and draw an arc above or below the line. Keeping the same setting, place the compass on $B$ and draw an arc intersecting the first one. Label this $Q$. Finally, draw a straight line directly through $Q$ and $P$.

Perpendicular from a Point NOT ON a Line

You can easily draw a line connecting a house to a road, but finding the absolute shortest path requires precise geometry. Drawing a perpendicular from an external point (a point hovering off the line) represents this exact shortest distance. First, place the compass on the external point $P$. Then, open the compass to a radius greater than the distance to the line. Draw an arc intersecting the base line at two points, $A$ and $B$. Next, place the compass on $A$ and draw an arc on the opposite side of the line from $P$. Keeping the same setting, place the compass on $B$ and draw an arc intersecting the one from the previous step. Label this $Q$. Finally, draw a straight line connecting $P$ and $Q$.

Constructing a 60° Angle

Did you know the secret to drawing a perfect $60^{\circ}$ angle is hiding inside a triangle? A 60 degree angle is constructed using the principle of an equilateral triangle, where all sides and interior angles are exactly equal. First, draw a base line and mark point $A$ as the vertex. Then, place the compass on $A$ and draw a large arc crossing the line and extending above it. Label the line intersection $B$. Next, without changing the arc radius (compass width), move the compass to $B$. Draw an arc intersecting the first large arc and label this $C$. Note that changing the compass width here will NOT create an equilateral triangle, ruining the angle. Finally, draw a straight line from $A$ through intersection $C$.