A geometrical construction is a proof made with lines and arcs. Instead of estimating where a point should go, the construction encodes a condition exactly: equal distance from two points, equal distance from two lines, a fixed side length, a fixed angle, or a combination of several constraints.
This twenty-eighth Secondary 4 Mathematics Learning Guide develops constructions, perpendicular bisectors, angle bisectors, triangle construction and constraint diagrams as one geometric reasoning system. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map.
Current syllabus connection: upper-secondary Geometry and Measurement includes construction of geometrical figures, perpendicular and angle bisectors, scale drawings and reasoning from stated geometric conditions. The examples below are original teaching material.
The perpendicular bisector means equal distance from two points
The perpendicular bisector of line segment AB crosses AB at its midpoint and at 90°. Every point on that bisector is equidistant from A and B.
If P lies on the perpendicular bisector of AB, then PA=PB.
The converse is also useful: if PA=PB, then P lies on the perpendicular bisector of AB.
Worked Example 1 | Locate points equidistant from two places
Two stations A and B are marked on a plan. A service point must be exactly the same distance from both stations. Where can it lie?
The service point can lie anywhere on the perpendicular bisector of AB. One condition gives a whole line of possible locations rather than one unique point.
How the perpendicular-bisector construction works
- Draw segment AB.
- With a compass radius greater than half of AB, draw arcs centred at A above and below the segment.
- Using the same radius, draw matching arcs centred at B.
- Join the two arc-intersection points.
The joining line is perpendicular to AB and passes through its midpoint. The equal compass radius builds pairs of points that are equally distant from A and B.
The angle bisector means equal distance from two intersecting lines
An angle bisector divides an angle into two equal angles. Points on an internal angle bisector are equidistant from the two sides of the angle, where distance to a line is measured perpendicularly.
An angle condition can become a distance condition.
Worked Example 2 | Equal distance from two roads
Two straight roads meet at point O. A kiosk must be placed inside the angle between them and be equally distant from both roads. Where can it lie?
It must lie on the internal angle bisector of the two roads.
How the angle-bisector construction works
- From the angle vertex O, draw an arc cutting both arms at points A and B.
- Using the same compass radius from A and B, draw arcs that meet inside the angle.
- Join O to the new arc-intersection point.
The construction creates two congruent triangles around the new line, forcing the two resulting angles to be equal.
Two constraints can produce one location
A single condition often gives a line or curve of possible points. A second independent condition can reduce that set to one or more intersections.
For example, “equidistant from A and B” gives the perpendicular bisector of AB. “Exactly 5 cm from C” gives a circle centred at C. Their intersection points satisfy both conditions.
Worked Example 3 | Intersect two geometric constraints
A point P must be equidistant from A and B and exactly 4 cm from A. Describe how to locate P.
Construct the perpendicular bisector of AB. Then draw a circle centred at A with radius 4 cm. Any intersection between the circle and the perpendicular bisector satisfies both conditions.
Depending on the geometry, there may be two, one or no intersections.
Triangle construction from three sides
If all three side lengths are known, one side can be drawn as a base. Arcs with radii equal to the other two side lengths locate the third vertex.
This is an exact geometric version of the fact that the third vertex must be simultaneously the required distance from both base endpoints.
Worked Example 4 | Construct a 5-6-8 triangle
Construct triangle ABC with AB=8 cm, AC=6 cm and BC=5 cm.
- Draw AB=8 cm.
- Draw an arc centred at A with radius 6 cm.
- Draw an arc centred at B with radius 5 cm.
- Choose an intersection as C and join AC and BC.
The second intersection would give the reflected congruent triangle on the other side of AB.
The triangle inequality predicts whether construction is possible
For three positive lengths to form a non-degenerate triangle, the sum of any two must exceed the third.
If the proposed lengths are 2 cm, 3 cm and 6 cm, construction fails because 2+3<6. The arcs cannot meet in a valid triangle.
Worked Example 5 | Predict construction failure
Can side lengths 4 cm, 7 cm and 12 cm form a triangle?
4+7=11<12, so no non-degenerate triangle is possible.
The geometric impossibility can be diagnosed before any drawing is attempted.
Constructing with two sides and an included angle
When two side lengths and their included angle are known, draw one side, construct the angle at the correct endpoint, then mark the second side length along that ray.
This fixes the triangle uniquely up to reflection when the data are valid.
Worked Example 6 | Two sides and included angle
Construct triangle ABC with AB=7 cm, AC=5 cm and ∠BAC=60°.
- Draw AB=7 cm.
- At A, construct a 60° ray.
- Mark C on the ray so AC=5 cm.
- Join C to B.
Scale construction changes representation, not geometry
If a plan uses 1 cm to represent 2 m, every real length must be converted by the same scale factor. Angles are preserved under ordinary scale enlargement or reduction.
A 10 m wall becomes 5 cm on the drawing. A 6 m wall becomes 3 cm. If their actual included angle is 70°, the drawing angle remains 70°.
Worked Example 7 | Scale drawing route
A triangular plot has two sides of 24 m and 30 m with included angle 50°. Draw it to scale 1 cm:6 m.
The scaled sides are 24/6=4 cm and 30/6=5 cm. Construct a 50° included angle between them. The third side can then be measured on the completed scale drawing if requested.
Perpendicular distance to a line is the shortest distance
When a point must be “within 3 cm of a straight boundary”, the distance is measured perpendicularly. The boundary of the allowed strip is formed by lines parallel to the original boundary at perpendicular distance 3 cm.
This turns verbal conditions into regions.
Worked Example 8 | Region satisfying a distance condition
A point must be no more than 2 cm from line l. Describe the allowed region.
Construct two lines parallel to l, each 2 cm away on opposite sides. The allowed region is the strip between and including those boundary lines.
Constructing a perpendicular from a point
A perpendicular from an external point P to a line l locates the shortest segment from P to that line. One construction route uses an arc from P to cut l at two points, then constructs the perpendicular bisector of the chord between them. That bisector passes through P.
Worked Example 9 | Why the method works
If an arc centred at P cuts line l at A and B, then PA=PB because both are radii. Therefore P lies on the perpendicular bisector of AB. Since AB lies along l, that bisector is perpendicular to l.
Constraint diagrams are easier when each condition is drawn separately
For a multi-condition location problem, translate one condition at a time:
- equal distance from A and B → perpendicular bisector;
- equal distance from two roads → angle bisector;
- fixed distance from A → circle centred at A;
- within a fixed distance of a line → parallel strip;
- closer to A than B → one side of the perpendicular bisector.
The valid region is the overlap of all required conditions.
Worked Example 10 | Combine three conditions
A point P must be closer to A than B, within 5 cm of C, and inside a given triangular region.
First construct the perpendicular bisector of AB and choose the half-plane containing A. Next draw a circle centred at C with radius 5 cm and take its interior. Finally intersect that region with the given triangle. The surviving overlap is the set of valid positions for P.
Accuracy matters because construction marks carry evidence
Construction arcs should be left visible when they are part of the required working. Erasing all arcs can remove the evidence showing how a bisector or perpendicular was produced.
A sharp pencil, stable compass width and careful ruler placement reduce accumulated error in scale drawings.
Common failure modes
| Error | Cause | Repair |
|---|---|---|
| Uses midpoint alone for equal distance from A and B | One point mistaken for full locus | Construct the whole perpendicular bisector |
| Draws angle bisector by visual estimation | Appearance substituted for construction | Use equal-radius arcs |
| Changes compass width between paired arcs | Equal-distance condition broken | Lock the radius until both arcs are drawn |
| Uses ordinary distance instead of perpendicular distance to a line | Distance definition unclear | Mark a 90° segment to the line |
| Applies one constraint and stops | Intersection logic missed | Draw each condition and keep only overlap |
| Erases all construction arcs | Working treated as clutter | Retain evidence required by the task |
Independent practice
- What line contains every point equidistant from points A and B?
- What line inside an angle contains points equidistant from its two sides?
- Describe how to construct a triangle with sides 4 cm, 6 cm and 7 cm.
- Can lengths 3 cm, 5 cm and 9 cm form a triangle? Explain.
- A scale drawing uses 1 cm:4 m. Convert actual lengths 28 m and 18 m into drawing lengths.
- Describe the region of points no more than 3 cm from a straight line l.
Explained answers
1. The perpendicular bisector of AB.
2. The internal angle bisector.
3. Draw the 7 cm base. Draw an arc of radius 4 cm from one endpoint and an arc of radius 6 cm from the other. Their intersection gives the third vertex.
4. No. 3+5=8<9, so the triangle inequality fails.
5. 28 m→7 cm; 18 m→4.5 cm.
6. The strip between two lines parallel to l and 3 cm away on either side, including the boundary lines.
Final thought
Construction questions become easier when each verbal condition is translated into a geometric object. Equal distance from points creates a perpendicular bisector. Equal distance from lines creates an angle bisector. Fixed distance creates a circle.
Draw the condition exactly. Then let the intersections reveal the answer.
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