Geometrical construction is not decorative drawing.
It is exact geometry produced from relationships such as equal distance, equal angles and perpendicularity.
A good construction works because every compass arc encodes an equality of distance. Those equalities force the line or point we want.
The direct answer
- A perpendicular bisector is the locus of points equidistant from two endpoints.
- An angle bisector is the locus of points equidistant from the two arms of an angle.
- Perpendiculars can be constructed by creating equal-distance points and joining them.
- Compass width matters because it preserves exact distance equality.
Perpendicular bisector of a line segment
Given segment AB:
- Set the compass radius to more than half of AB.
- With centre A, draw arcs above and below AB.
- Without changing the compass width, repeat from B.
- Let the arc intersections be P and Q.
- Draw line PQ.
PQ is the perpendicular bisector of AB.
It cuts AB in half and meets it at 90°.
Why the perpendicular-bisector construction works
Because P was created with the same compass radius from A and B:
PA=PB.
Likewise:
QA=QB.
So P and Q are both equidistant from A and B.
The set of all points equidistant from A and B is the perpendicular bisector of AB.
The construction succeeds because the compass does not guess the midpoint. It creates two points whose equal-distance conditions force the midpoint line.
Finding the midpoint as a by-product
Let M be the intersection of PQ and AB.
Then:
AM=MB.
So the perpendicular-bisector construction simultaneously produces:
- the midpoint of AB;
- a line perpendicular to AB.
Angle bisector
Given ∠ABC:
- With centre B, draw an arc cutting BA and BC at D and E.
- Without changing the next chosen compass width, draw arcs from D and E so they intersect at F inside the angle.
- Draw BF.
BF bisects ∠ABC.
Therefore:
∠ABF=∠FBC.
Why the angle-bisector construction works
Because the first arc was centred at B:
BD=BE.
Because the second pair of arcs used equal radius:
DF=EF.
BF is common.
Therefore triangles BDF and BEF are congruent by SSS.
So their angles at B are equal.
The angle bisector is a congruence construction: equal compass radii build two matching triangles around the desired bisecting line.
Perpendicular from a point on a line
Suppose P lies on line l and we want a perpendicular through P.
- With centre P, draw an arc cutting l at A and B.
- Using equal compass radius from A and B, draw arcs intersecting at Q away from the line.
- Draw PQ.
PQ is perpendicular to l.
The reason is the same equal-distance structure as the perpendicular bisector: P and Q lie on the perpendicular bisector of AB.
Perpendicular from a point outside a line
Suppose P lies outside line l.
- With centre P, draw an arc large enough to cross l at A and B.
- From A and B, draw equal-radius arcs intersecting at Q on the opposite side of l from P.
- Draw PQ.
Since PA=PB and QA=QB, both P and Q lie on the perpendicular bisector of AB.
Therefore PQ is perpendicular to AB and hence to line l.
Why compass width must be preserved when required
If two arcs are intended to create equal distances, changing the compass radius destroys the equality.
A construction diagram may still look plausible, but its proof no longer works.
That is why construction marks and arcs are part of the mathematics, not rough working to erase immediately.
The perpendicular bisector as a locus
Every point on the perpendicular bisector of AB is equidistant from A and B.
Conversely, every point equidistant from A and B lies on that perpendicular bisector.
This two-way statement makes the construction especially useful in loci problems.
The angle bisector as a locus
Every point on an angle bisector is equidistant from the two arms of the angle, where distance to a line means perpendicular distance.
Conversely, a point inside the angle equidistant from the two arms lies on the angle bisector.
This creates the bridge from construction into region and locus questions.
Constructing the circumcentre of a triangle
The circumcentre is the point equidistant from all three vertices.
Construct perpendicular bisectors of two sides.
Their intersection O satisfies:
OA=OB and OB=OC.
Therefore:
OA=OB=OC.
A circle centred at O through one vertex passes through all three.
Constructing the incentre of a triangle
The incentre is the point equidistant from the three sides of a triangle.
Construct angle bisectors of two triangle angles.
Their intersection lies at equal perpendicular distance from the corresponding side pairs and therefore from all three sides.
This point is the centre of the incircle.
Perpendicular bisectors locate a point equally far from vertices. Angle bisectors locate a point equally far from sides. The construction chosen should match the kind of equality the problem asks for.
Constructing 60° and 30° angles
An equilateral triangle can be built by intersecting equal-radius arcs from the endpoints of a segment.
All three sides are equal, so all angles are 60°.
Bisecting a 60° angle produces 30°.
This shows how a small set of constructions can generate further exact angles.
Constructing 90° and 45° angles
A perpendicular construction gives 90°.
Bisect that right angle to obtain 45°.
Again, exactness comes from geometric relationships rather than protractor measurement.
Construction versus measurement
A ruler and protractor can estimate or measure a drawing.
A classical construction establishes a relationship exactly within Euclidean geometry.
The two tasks should not be confused.
If the instruction is “construct”, numerical measurement should not replace the required arcs and lines.
Why construction arcs should remain visible
The arcs show how the result was generated.
Without them, a correct-looking line may be indistinguishable from a guessed line.
In assessed work, construction evidence demonstrates the method.
A practical construction routine
- Read what relationship is required: equal distance, midpoint, perpendicularity or equal angles.
- Select the construction whose locus matches that relationship.
- Keep required compass widths unchanged.
- Draw arcs clearly enough to show intersections.
- Join only the points forced by the construction.
- State or mark the resulting property.
Common misconception 1: perpendicular bisector means any perpendicular line
It must also pass through the midpoint of the segment.
Common misconception 2: angle bisector halves the side opposite the angle
An angle bisector halves the angle. It does not generally bisect the opposite side.
Common misconception 3: arcs are merely rough guides
Their equal radii encode the equal-distance facts that prove the construction.
Common misconception 4: measuring 90° with a protractor is the same as constructing a perpendicular
Measurement and construction are different mathematical tasks.
A diagnostic ladder
- Can the learner construct a perpendicular bisector?
- Can the learner explain its equal-distance locus?
- Can the learner construct an angle bisector?
- Can the learner prove the angle halves by congruence?
- Can the learner construct a perpendicular through a point on a line?
- Can the learner construct a perpendicular from an external point?
- Can the learner locate a circumcentre?
- Can the learner locate an incentre?
- Can the learner preserve construction evidence instead of replacing it with measurement?
The deeper lesson: constructions turn equal distances into exact structure
Compass arcs do not know about “midpoint” or “angle bisector”.
They only preserve distance.
Yet carefully chosen equal-distance conditions force midpoint, perpendicularity, symmetry and equal angles to appear.
Geometrical construction is powerful because exact structure emerges from a very small toolkit: equal radii create loci, loci intersect at forced points, and those points reveal the perpendiculars, bisectors and centres the problem asks for.