Thus, in one demonstration, we have to show the equality of three different kinds of magnitudes, lines, angles, and area or surface. Have we any test of equality which will apply to them all? Axiom 8 tells us that magnitudes which coincide are equal, and, further, it was pointed out in a note that the test of coincidence could be applied to the three magnitudes in question. But how are we to apply this test? From the same note we gather that one triangle must be set upon or applied to the other. We will then imagine triangle ABC to be applied to DEF. Now, since in the two triangles two sides and the angle between them are given equal, we know that these equal parts will coincide. What results can we observe from this ? First, since the equal sides coincide the bases must coincide, for the extremities of the bases are also the extremities of the sides which coincide, and if the extreme points of two straight lines coincide the whole lines must do so also, for if not two straight lines would enclose a space.

Next, since the bases coincide, all the sides of the triangles coincide. Therefore the parts contained by them must coincide; these are (1) the area, i.e. the whole space included by the sides; (2) the two remaining angles in each triangle contained by the base and one of each of the equal sides.

Finally, the parts which coincide must, as we have seen above (ax. 8), be equal.

Therefore the triangles ABC, DEF, are equal as to their sides, their angles, and their area-Which was to be proved.

SUMMARY OF PROPOSITION IV., THEOREM 1.

Given: In two triangles two sides and the included angle equal.

To be demonstrated: That the triangles are equal in all respects.

Cons.-Nil.

Proof.-Apply one triangle to

the other.

Those parts which are given equal will coincide.

D

Да

B

C E

the equal sides coincide at all points, the bases must coincide (post. 5), and are equal (ax. 8).

all the sides coincide, the parts included by them, i.e. the areas and remainings, coincide and are equal (ax. 8).

The method of superposition, as it is called, is here made use of. It consists in supposing one figure to be laid upon another for the purpose of comparison, and depends upon the eighth axiom. It is made use of sparingly by Euclid, but more often by later geometers of the Alexandrian school and by modern writers.

The third sides of the triangles in this proposition are spoken of as 'bases.' This name may be applied to any side of a triangle to distinguish it from the other two, except in an isosceles triangle, when it always refers to the unequal side. The point opposite the base, at which the other two sides of a triangle meet, is called the 'vertex,' and the angle at this point the 'vertical angle.'

Prop. V. Before beginning the next theorem, I wish you to notice that from Proposition IV. we learn (1) that triangles are equal in all respects, if in each triangle two sides and the angle between them are severally equal; therefore if we want to show that any two triangles are equal we must prove that two sides and the included angle in one triangle are equal to the corresponding parts in the other. (2) That the triangle affords a means of comparing angles, for by showing the equality of the triangles in the last proposition we were able to prove the equality of two pairs of angles.

You will see that we can include, or shut in, any angle in a triangle, by drawing a straight line meeting the lines containing the angle, or the arms of the angle as they are called. With this preface I hope that we shall not only be able to understand, but to work out for ourselves the famous fifth proposition.

We are required to show :

(1) That the angles at the base of an isosceles triangle are equal.

(2) That if the equal sides are produced the angles on the other side of the base are equal.

Draw an isosceles triangle ABC of which side AB = side AC we have to show (1) that angle ABC = angle ACB.

Produce the equal sides to D and E we have to show (2) that angle DBC is = to angle ECB.

We will first endeavour to show that D

B

C

E

angle DBC is equal to angle ECB. We have said that in order to compare angles we must include them in triangles. Now look at the figure and you will see that we can construct a triangle containing angle DBC by drawing a line from C to any point F in BD. In the same way angle BCE can be included in a triangle by drawing a line from B to a point G in CE. But in order to show that the angles are equal we must show that the triangles containing them are equal; therefore in constructing the triangles we must do what we can to make them equal, so let us take point G, so that CG (a side of one triangle) is equal to BF a side of the other.

We will pause to consider our figure,

we have now :

(1) The given triangle ABC.

(2) A pair of constructed triangles

FBC, GCB.

(3) A pair of triangles AFC and AGB formed by adding the given triangle to each of the constructed ones.

D

F

A

You will see, too, that as the triangles overlap each other, some of the sides and some of the angles are common to two or even three triangles. Thus, taking the angles first, we have the angle at A common to three triangles, the given triangle and the larger pair; the angles at F and G are each common to one of the larger and one of the smaller con

structed triangles. Similarly, taking the sides, we have BC common to three triangles; FC and BG, each of them common to two, as also BA and AC.

Let us proceed to compare the smaller pair of triangles, for if we can show that they are equal, it follows that 'the angles on the other side of the base' are equal.

In the triangles FBC, BCG:

we have side BF
side BC

CG (by construction)
BC (common to both).

We know nothing of the size of any of the angles, but the angles at F and G, as also sides FC and BG, are, as we have said, common to the larger and smaller pair of triangles. Let us then compare the larger triangles, and if they are equal we can retrace our steps and show that the smaller pair are. In triangles FAC, GAB, we have

F

B

G

side AB = side AC (sides of the isosceles

triangle)

side AF side AG, since their parts are equal (AB = AC and BF = CG cons.)

The angle at A is common to both, and is contained by the equal sides, therefore by Proposition IV. these triangles are equal in every respect. We return to the smaller pair of triangles; we can now assert that

E

side BF = side CG (cons.)

=

side FC side BG (since they are sides of larger pair of triangles shown to be equal by I. 4),

and the angles at F and G, included by the equal sides, are equal (since they are also angles of larger pair of triangles); these triangles therefore are equal in all respects (I. 4), and angles DBC and ECB, the angles on the other side of the base, are equal.

We must now show that the angles at the base are equal, angle ABC to angle ACB. Glancing at the figure again you will see that the angles ABG and ACF are each made

up

of

one of the 'angles at the base' and one of the angles of the smaller pair of equal triangles, or, to put it more clearly,

And we know that angles ABG and ACF are equal, since they are similar angles of the large pair of triangles. Also angles CBG and BCF are equal, since they are similar angles of the smaller pair of triangles. Then subtracting the equal parts from the equal wholes we have equal remainders (ax. 3), that is the angle ACB is equal to the angle ABC, and these are the angles at the base.

SUMMARY OF PROPOSITION V., THEOREM 2.

The angles at the base of an isosceles triangle are equal, and if the equal sides be produced, the angles on the other side of the base are equal.

Cons.-In BD take any point F. From

CE cut off CG BF (I. 3), join FC, GB.

=

Proof.-By I. 4 prove (1) the larger pair of triangles equal, (2) the smaller pair of triangles equal.

Since the smaller As contain the s on the other side of the base,' these angles are equal.

Subtract from the Ls of the larger triangles, the 8 of the smaller which form a part of them, and the remainders, 'thes at the base,' are equal (ax. 3).

From this proposition it is plain that all the three angles of an equilateral triangle are equal.

A truth which follows immediately from the proof of any theorem is called a corollary.'

Proposition V. is of very ancient origin, and is ascribed to the philosopher Thales, one of the seven wise men of Greece.