Geometric Constructions

What the tools are allowed to do

An unmarked straightedge draws a line through two distinct known points. A compass draws a circle with a known centre and an established radius. Intersections of previously constructed lines and circles produce new points.

A protractor, a marked scale and visual estimation are useful measuring tools, but they are not part of a strict compass-and-straightedge construction. Do not label a line “perpendicular” solely because it looks vertical on the page.

A construction should state its input, its steps and its justification. Keeping the light construction arcs in the final drawing records how the result was obtained.

Constructing a perpendicular bisector

Given a non-zero segment $AB$, choose a compass radius larger than half its length. Draw equal-radius circles centred at $A$ and $B$, meeting at two points $P$ and $Q$. Draw line $PQ$. It is the perpendicular bisector of $AB$.

Why it works

Both $P$ and $Q$ are equidistant from $A$ and $B$ because they lie on equal-radius circles. Comparing $\triangle APQ$ and $\triangle BPQ$ gives SSS congruence. Their matched angles at $P$ are equal. If $M=PQ\cap AB$, triangles $APM$ and $BPM$ are congruent by SAS, so $AM=BM$ and their adjacent angles at $M$ are equal. Those adjacent angles total $180^\circ$, so each is $90^\circ$.

For $A(-3,0)$, $B(3,0)$ and radius $5$, the two circle equations are

Subtracting the equations gives $12x=0$. Substituting $x=0$ then gives $y^2=16$, so the intersections are $P(0,4)$ and $Q(0,-4)$. The line $x=0$ bisects the segment at $(0,0)$. A radius exactly $3$ would produce only one circle intersection, insufficient to define $PQ$; a smaller radius produces none.

Constructing an angle bisector

Given an angle strictly between $0^\circ$ and $180^\circ$ with vertex $O$, draw one circle centred at $O$ meeting the rays at $X$ and $Y$. Then draw equal-radius circles centred at $X$ and $Y$, using a radius large enough for two intersections. Choose their intersection $P$ inside the angle, distinct from $O$. Join $O$ to $P$.

The triangles $OXP$ and $OYP$ have $OX=OY$, $XP=YP$ and common side $OP$. SSS congruence makes their angles at $O$ equal, so $OP$ bisects the angle. The equal compass settings are the proof’s essential information.

For a right angle with $X(4,0)$ and $Y(0,4)$, radius-$4$ circles centred at those points meet at $O(0,0)$ and $P(4,4)$. The ray $OP$ is $y=x$ and splits the $90^\circ$ angle into two $45^\circ$ angles. Choosing $O$ as the new point would not define a new ray.

The construction and its justification correspond to Euclid I.9.

Drawing perpendiculars and parallels

To construct a perpendicular from a point $P$ outside a line, draw a circle centred at $P$ that crosses the line twice, at $X,Y$. Construct the perpendicular bisector of $XY$. Since $PX=PY$, the point $P$ lies on that bisector, so the resulting line passes through $P$ and is perpendicular to the original line.

For example, a radius-$5$ circle centred at $P(1,4)$ meets the $x$-axis at $X(-2,0)$ and $Y(4,0)$. Their midpoint is $(1,0)$, directly below $P$. The construction returns the perpendicular line $x=1$.

To construct a perpendicular at a point on a line, mark two points on the line at equal distance from that point and construct their perpendicular bisector. To obtain a parallel through an external point, construct a perpendicular through that point and then a second perpendicular to the first at the same point. Two lines perpendicular to the same line are parallel in the plane.

Constructing a triangle from three sides

Suppose the required sides are $AB=6\text{ cm}$, $AC=5\text{ cm}$ and $BC=4\text{ cm}$. Draw $AB$. Draw a circle of radius $5$ centred at $A$ and a circle of radius $4$ centred at $B$. Either intersection can serve as $C$.

Both resulting triangles have the required sides, and the two choices are reflections in $AB$, hence congruent. The circles intersect twice because the lengths satisfy the strict triangle inequalities. If the longest side equalled the sum of the others, the circles would touch and the “triangle” would be degenerate.

With coordinates $A(0,0)$ and $B(6,0)$, the upper intersection is $C(15/4,5\sqrt7/4)$. This coordinate result is a verification of the construction, not a requirement to calculate coordinates before drawing it.

Constructing a circumcircle

Three non-collinear points determine one circle. Construct the perpendicular bisectors of two sides of their triangle and call their intersection $O$. The first bisector gives equal distances to its two endpoints, and the second gives equality to the third point. Therefore $O$ is equally distant from all three vertices.

Draw the circle centred at $O$ through any vertex. It passes through the other two. For the right triangle with vertices $(0,0)$, $(6,0)$ and $(0,8)$, the centre is $(3,4)$ and the radius is $5$, the midpoint and half-length of the hypotenuse.

An obtuse triangle’s circumcentre lies outside the triangle. That is not a construction failure. Collinear input points, by contrast, do not determine an ordinary circle through all three.

Construction checks and common errors

Check that a line passes through the intended intersections and that circles that should have equal radii used the same compass opening. Two arcs that fail to intersect do not justify extending an imagined meeting point by eye; the chosen radius or construction step must be corrected.

A diagram can show an intended method even when hand-drawn tolerances are imperfect. The proof must still identify why the ideal intersections satisfy the target condition. “It measures about $90^\circ$” is a measurement check, not the explanation of a perpendicular construction.

Related topics

Congruence explains why equal-radius constructions work. Proof in Geometry turns construction steps into justified arguments, and Reflection uses perpendicular bisectors to locate mirror lines.