Constructions, Loci and Congruence

This lesson brings together three practical, mark-rich parts of OCR GCSE Mathematics (J560): accurate constructions with ruler and compasses, loci (the set of points obeying a rule), and the closely linked ideas of congruence and similarity. Constructions reward precision and leaving your compass arcs visible; loci reward translating a worded condition into a region or path; and congruence questions reward quoting the correct condition (SSS, SAS, ASA or RHS) as a reason. Performing a construction or a calculation is AO1; reasoning about why two triangles are congruent is AO2; and applying loci or similarity to a problem is AO3.

Key Vocabulary

Term	Meaning
Perpendicular bisector	A line cutting another at $90^{\circ}$90∘ through its midpoint
Angle bisector	A line cutting an angle into two equal halves
Locus (plural loci)	The set of all points satisfying a given rule
Equidistant	The same distance from two or more things
Congruent	Identical in shape and size
Similar	Same shape, different size (sides in proportion)
SSS / SAS / ASA / RHS	The four conditions that prove triangles congruent

Ruler-and-Compass Constructions

A construction is an accurate geometric drawing made using only two tools: a straight edge for drawing lines and a pair of compasses for drawing arcs. For the angle constructions in particular you must not use a protractor — the whole point is to produce the angle exactly using arcs rather than by measuring. The single most important habit, and the one that most affects marks, is to leave all of your construction arcs visible on the page. Examiners award the method marks for the arcs themselves, because the arcs are the evidence that you used a valid construction rather than guessing or measuring; a perfect final line with no arcs typically scores almost nothing. Use a sharp pencil, keep the compass setting fixed when the method requires it, and draw arcs that are long enough to cross clearly. The two foundational constructions, from which most others are built, are the perpendicular bisector of a line segment and the bisector of an angle. To bisect a line $AB$AB, open the compasses to a radius bigger than half the length of $AB$AB, place the point on $A$A and draw an arc above and below the line, then — without changing the radius at all — place the point on $B$B and draw a second pair of arcs. The arcs intersect at two points, one above the line and one below; joining these two points with the straight edge gives the perpendicular bisector, which cuts $AB$AB exactly in half and at a right angle. The figure shows this construction with the arcs left in place.

Worked Example 1

Describe how to construct the perpendicular bisector of a line segment $PQ$PQ.

Set the compasses to more than half of $PQ$PQ. With the point on $P$P, draw an arc above and below the line; without changing the radius, repeat with the point on $Q$Q. The arcs cross at two points; join them with a straight line. That line is the perpendicular bisector, cutting $PQ$PQ at $90^{\circ}$90∘ through its midpoint.

Answer: join the two arc intersections.

Worked Example 2

Describe how to bisect an angle $ABC$ABC.

With the compass point on the vertex $B$B, draw an arc crossing both arms of the angle. From each of those two crossing points, draw an arc in the interior; these two interior arcs cross at one point. Join $B$B to that point: the join bisects the angle into two equal halves.

Answer: join the vertex to the interior arc intersection.

Common error: rubbing out the construction arcs — leave them visible, because they earn the method marks.

Worked Example 3

Describe how to construct a $60^{\circ}$60∘ angle at a point on a line.

Mark a point $X$X on the line. With the compasses on $X$X, draw an arc cutting the line at $Y$Y. Without changing the radius, place the point on $Y$Y and draw a second arc crossing the first at $Z$Z. Join $X$X to $Z$Z: angle $ZXY$ZXY is exactly $60^{\circ}$60∘, because $XYZ$XYZ is an equilateral triangle.

Answer: the equilateral-triangle construction gives $60^{\circ}$60∘.

Loci

A locus (the plural is loci) is the set of all points that satisfy a given geometric condition. The word sounds technical, but the idea is intuitive: it is simply the path or region traced out by every point that obeys some rule. The key skill is turning a worded rule into an accurate drawing, and to do that it helps enormously to know the four standard loci that cover almost every exam question. First, the locus of points a fixed distance from a single point is a circle centred on that point. Second, the locus of points a fixed distance from a straight line is a pair of parallel lines on either side, joined by semicircular ends if the line is a segment — often described as a "racetrack" shape. Third, the locus of points equidistant from two points is the perpendicular bisector of the line segment joining them, which is exactly why the perpendicular-bisector construction is so important. Fourth, the locus of points equidistant from two lines is the angle bisector of the angle between them. Many questions combine two or more of these conditions, and then the answer is the region where all the individual loci overlap, which you show by shading. Always read carefully whether the question wants a path — a line or curve — or a region, an area to be shaded. The figure shows the locus of points a fixed distance $r$r from a point $C$C.

Worked Example 4

A goat is tied to a post by a rope of length $4$4 m. Describe the locus of points the goat can reach, and state its boundary.

The goat can reach any point up to $4$4 m from the post, so the locus is a circle of radius $4$4 m centred on the post, together with its interior. The boundary is the circle itself.

Answer: a circular region of radius $4$4 m about the post.

Worked Example 5

Describe the locus of all points equidistant from two fixed points $A$A and $B$B.

Every point the same distance from $A$A and from $B$B lies on the perpendicular bisector of $AB$AB, so the locus is that perpendicular bisector.

Answer: the perpendicular bisector of $AB$AB.

Worked Example 6

A point must be within $3$3 cm of a point $P$P and closer to line $\ell_1$ℓ1​ than to line $\ell_2$ℓ2​. Describe the region.

The first condition gives the inside of a circle of radius $3$3 cm about $P$P; the second gives the side of the angle bisector of $\ell_1$ℓ1​ and $\ell_2$ℓ2​ nearer $\ell_1$ℓ1​. The region is the overlap of these two — the part of the circle's interior on the $\ell_1$ℓ1​ side of the bisector.

Answer: the intersection of the disc and the half-plane.

Common error: drawing the boundaries but not shading the correct overlapping region that satisfies both conditions.

Congruence