Ruler and compass
Tonight's idea
The Greek rules: an unmarked ruler, which draws the line through two points, and a compass, which draws a circle with a chosen centre through a chosen point. No measuring. Every step then has a reason, almost always a pair of congruent triangles.
Two tests do the work. SSS: two triangles with the same three side lengths are congruent, so their angles match too. SAS: two sides and the angle between them fix the triangle.
Bisect a segment \(AB\). Draw circles centred at \(A\) and at \(B\) with the same radius \(r\), bigger than half of \(AB\). They meet at \(P\) and \(Q\). Draw \(PQ\); it crosses \(AB\) at \(M\).
Why it works. Triangles \(PAQ\) and \(PBQ\) share \(PQ\) and have \(PA = PB = QA = QB = r\): SSS, so \(\angle APM = \angle BPM\). Then triangles \(APM\) and \(BPM\) have \(AP = BP\), \(PM\) shared, and equal angles between them: SAS. So \(AM = BM\), and the two angles at \(M\) are equal and add to \(180°\): each is \(90°\).
Watch one: bisecting an angle
Put the compass point on the vertex \(V\) and draw an arc cutting both arms, at \(A\) and \(B\). Then, using one radius for both, draw arcs from \(A\) and from \(B\); they meet at \(P\). Draw \(VP\).
Reason: \(VA = VB\) because both are radii of the first arc, \(AP = BP\) because the second and third arcs had equal radii, and \(VP\) is shared. SSS, so \(\angle AVP = \angle BVP\). The compass never measured anything; it only copied a length.
Your turn
- A point \(P\) lies off a line. Drop the perpendicular from \(P\) to the line: give the steps (two circles and one line is enough), then explain why the line you drew meets the first line at right angles. The idea section has already proved what you need.
- Construct a \(60°\) angle at a point \(A\) on a line, using the fact that a triangle with three equal sides has three equal angles. Then say how to get \(30°\) and \(15°\).
- Copy an angle: given an angle at \(V\) and a ray starting at \(W\), construct the same angle at \(W\). Give the steps and name the congruence test that proves the two angles are equal.
- Through a point \(P\) not on line \(\ell\), construct the line parallel to \(\ell\). One method: pick \(A\) on \(\ell\); draw the circle centred \(A\) through \(P\), meeting \(\ell\) at \(B\); draw circles of the same radius centred \(B\) and \(P\), meeting at \(Q\) on the far side from \(A\). Explain why \(PQ \parallel \ell\), using one SSS pair and the fact from a11 that equal alternate angles mean parallel lines.
Stretch: the lost centre
Someone drew a circle and lost the centre. Find it again with ruler and compass, and say why your construction must land on the centre. Then: the Greeks could bisect any angle, as above. They could not find a way to cut an angle into three equal parts. Why do you think that is harder?
Answers
- Mark any two points \(A\) and \(B\) on the line. Draw the circle centred \(A\) through \(P\) and the circle centred \(B\) through \(P\); they meet again at \(Q\), on the other side of the line. Draw \(PQ\). Now \(AP = AQ\) and \(BP = BQ\) (radii), so \(A\) and \(B\) are each equidistant from \(P\) and \(Q\). That is all the proof in the idea section used, with the two pairs of letters swapped (it never needed the two radii to be equal to each other). So the line \(AB\) is the perpendicular bisector of \(PQ\): \(PQ\) meets the line at right angles, and it passes through \(P\) by construction.
- Draw a circle centred \(A\) meeting the line at \(B\); draw the circle centred \(B\) with the same radius; the circles meet at \(C\). \(AB = BC = CA\), so triangle \(ABC\) has three equal angles which add to \(180°\): \(\angle CAB = 60°\). Bisect it (example above) for \(30°\), and bisect again for \(15°\).
- Draw an arc centred \(V\) cutting the arms at \(A\) and \(B\). Draw the same arc centred \(W\), cutting the ray at \(A'\). Set the compass to the distance \(AB\) and draw an arc from \(A'\); it cuts the arc at \(W\) in \(B'\). Draw \(WB'\). Triangles \(VAB\) and \(WA'B'\) have \(VA = WA'\), \(VB = WB'\) (same radius) and \(AB = A'B'\) (copied), so they are congruent by SSS and \(\angle AVB = \angle A'WB'\).
- All of \(AP, AB, BQ, PQ\) equal the first radius, so \(ABQP\) is a rhombus. Draw the diagonal \(AQ\). Triangles \(ABQ\) and \(QPA\) have the same three sides (\(AB = QP\), \(BQ = PA\), \(AQ\) shared): SSS, so \(\angle BAQ = \angle PQA\). These are alternate angles made by the line \(AQ\) crossing \(\ell\) and \(PQ\), so \(PQ \parallel \ell\). (Another route: drop the perpendicular from \(P\) to \(\ell\), then construct the perpendicular to that line at \(P\).)
Stretch. Mark any three points \(A, B, C\) on the circle and construct the perpendicular bisectors of \(AB\) and \(BC\). They meet at the centre: every point of the first bisector is equidistant from \(A\) and \(B\), every point of the second from \(B\) and \(C\), and the centre is the one point equidistant from all three. Bisecting an angle only ever copies lengths, which is what a compass does. Trisecting needs a length that no chain of circles and lines can reach from the starting points; that this is impossible for a general angle was proved in 1837, more than two thousand years after the question was first asked.