triangles ADC and BDC were shown to be equal, the angles ADC and BDC are equal; and therefore, according to the definition just quoted, they must be right angles, and DC is drawn perpendicular to AB. Then supposing AB to be the given line, and D the given point, DC is the perpendicular required. But AB in the last proposition was bisected at D, and the given point is not necessarily the point of bisection. Produce AB in one or both directions, and you will see that the construction still holds good. It remains for us to apply it to the problem we are engaged upon.

Let AB be the given line, and C the given point.

Since the given point bisects the base of the equilateral triangle we must mark off equal parts, CD, CE, on each side

F

A E

C

D

B

E

of the point C (I. 3). DE will then form the required base bisected at C, on it describe the equilateral triangle DFE. A line joining the vertex of this triangle with the given point will be the perpendicular required.

But the given point may be at one extremity of AB, as at A, in this case we must produce the line any distance beyond A, marking off on the other side of it a part equal to the line produced.

A

A

D

B

One case alone is noticed by Euclid.

SUMMARY OF PROPOSITION XI., PROBLEM 6.

To draw a straight line at right angles to a given straight line from a given point within it.

E

D B

Cons. Cut off equal parts DC, EC, on each side of the given point (I. 3). On DE (DC + EC) describe an equilat. DFE (I. 1). Join FC.

Proof.- DFC = ^ EFC (I. 8)

=

.. 4 DCF 2 ECF (and they are adjacent Ls)
.. FC is perpendicular to AB (def. 10).

Proposition 12 is closely related to the last; it requires us to draw a line at right angles to another from a given point without it. Euclid further specifies that the given line shall be of unlimited length; if not, the point might be so placed that it would be impossible to let fall a perpendicular from it to the given line, a b. In the figure of I. 11, supposing

F to be the given point, FC is the perpendicular required.

The perpendicular was obtained by joining the vertex of an equilateral triangle with the point bisecting its base. This triangle was described in order to find two equal lines drawn from fixed points within the given line to a point without it not found. The condition is now reversed, the equal lines must be drawn from the fixed point without the line, therefore we cannot employ Prop. I.

G

C

But equal lines can be obtained by means of the circle, the given point C from which the lines are to be drawn will form the centre, what will be the distance ? The extremities of the equal lines must lie in AB, hence the circle must cut AB in two points. AB is of unlimited length, therefore, with C as centre and any point D on the other side of AB as distance, a circle can be drawn cutting AB (produced if necessary) in two points E and F. To these points draw the radii CE and CF. The remainder of the construction is obvious, bisect FE in G, join GC, which is the perpendicular required.

D

SUMMARY OF PROPOSITION XII., PROBLEM 7.

E

To draw a perpendicular to a given line from a given point without it.

Cons. With given point as centre

and any point on the other side of the line as radius, describe a circle.

Bisect that part of the line which the circle intercepts, join given point to point of bisection and to the two points at which the circle cuts AB.

G

E

B

D

Proof. The two similar As are equal (I. 8);

.. the adjacent s FGC, EGC are equal
.. GC is perpendicular to AB.

This proposition concludes the second group of problems, in which we have learnt :

:

To bisect (a) an angle,

(B) a finite straight line.

To draw a perpendicular to a line from a point (a) within it, (B) without it.

The solution of the last proposition chiefly depended on finding two points equidistant from the given point C, and lying in AB. We observed that the circumference of a circle drawn round C would furnish an infinite number of points, satisfying the first condition, and any points in AB of course satisfy the second. The points of intersection of AB and the circle were the only ones which would satisfy both conditions.

This is called the method of loci.

In plane geometry a locus is a line, curved or straight, every point of which satisfies a certain condition.

The intersection of loci is often a ready means of determining the position of a point necessary for the solution of a problem, since straight lines will only intersect at one point and circles either with each other, or with straight lines at two.

I. 1 is an example of the intersection of loci, for a circle described round A at the distance AB is the locus of all points at that distance from A. Similarly, a circle of which B is the centre is the locus of all points at the required distance from B. One of the points at which these circles cut, i.e. the intersection of the loci, must form the vertex of the required triangle.

The method of loci is ascribed to the Platonic school.

LECTURE VII.

PROPOSITIONS 13-15.

A GROUP of three theorems follows, the demonstration of all of which chiefly depends on the fact that right angles are equal.

Prop. XIII. teaches us that 'the angles which one straight line makes with another on the same side of it either are, or are together equal to two right angles.

This theorem is almost self-evident, for we know by def. 10 that if the angles are equal they are right angles, and that in this case one line is perpendicular to the other.

But instead of being perpendicular let us suppose that one line is inclined a little to the other, is it not evident that while one of the angles is made larger than a right angle, the other is made proportionately smaller, and that the two angles taken together are still equal to two right angles?

It remains to show this formally. In the lesson on the definitions we noticed that Euclid did not measure his angles exactly, but took the right angle as a standard of comparison, speaking of other angles as being greater or less than a right angle. Let us apply this test here.

Let AB be a line falling upon CD so that the angles are unequal in size. To compare these angles with right angles we must draw a line perpendicular to CD from point B (I. 11).

The figure is now divided into two given angles ABC, ABD, two right angles EBC, EBD, three angles EBC, EBA, ABD. The angles EBA, ABD

E

together make up and are therefore

D

B

equal to angle EBD, to complete the figure we have angle EBC.

Therefore EBC + 2 EBD = ¿ EBC + ▲ ABE + ▲ ABD, i.e. the two right angles the three angles.

=

Again, angle ABC is made up of and is therefore equal to angles EBC and ABE, to complete the figure we must add ▲ ABD.

Then ▲ ABC + 2 ABD = 2 EBC + 2 ABE + ▲ ABD, i.e. the two given angles the three angles.

=

Therefore the two given angles = the two right angles (ax. 2).

SUMMARY OF PROPOSITION XIII., THEOREM 6.

The angles which one straight line makes with another straight line on one side of it, either are two right angles or are together equal to two right angles.

PROP. XIV.

The enunciation of the next theorem is 'If at a point in a straight line two other straight lines on the opposite sides of it make the adjacent angles together equal to two right angles, these lines shall be in one and the same straight line.'

The meaning of this will become apparent by an example. Let AB be any straight line, and let C and D be points one on each side of the point B, then supposing straight lines to be