DF being equal to DC, we may be led to proceed in one of two ways:—

First, we might notice that our construction made DG equal to DC; so that DF is equal to D G, therefore the angle DFG equal to the angle D G F (Euc. I., 5); therefore the angle B F G greater than the angle BG F; and BG greater than BF (Euc. I., 19); that is, BD, DC, together greater than B A, A C together.

Or, we might notice that since D F is equal to DC, BD and DF are together equal to BD and DC together; but BD and D F are together greater than BF (Euc. I., 20); therefore BD and CD are together greater than BA and A C together.

If we had followed the first of these courses, we should still scarcely fail to notice afterwards that the second is an available and a better solution.

We proceed, then, to run over such steps of the above work as are necessary to the proof of the proposition. In doing this we notice that a property of the third book has been made use of in proving that CF is at right angles to AH. We will assume that the proposition has been given as a deduction from the first book. Then, although the student might mentally have followed the course we have adopted, it would be well for him to modify the proof so as to avoid the use of Book III. This is easily done. The student sees at once that the proof involves the equality (in all respects) of the triangles AHF, AH C. He had the angle A F H equal to the angle

ACH, AF equal to A C, and AH common. This is not quite sufficient (so far as Euclid's treatment of triangles extends). But it is easy to supplement these data by establishing the equality of the angles FAH, HAC, these angles being respectively equal to the equal angles A B C, ACB (Euc. I., 29). Hence the triangles H AF, HAC are equal in all respects.

The construction and proof of the proposition we are dealing with run, therefore, thus:

Produce BA to F, making AF equal to AC. Join DF, DC, and let A D, produced if necessary, meet FC in H. Then the angle FAH is equal to the interior angle ABC (Euc. I., 29). But ABC is equal to ACB (Euc. I., 5), and ACB to CA H (Euc. I., 29). Therefore the angle FA H is equal to the angle CAH. Also FA, A H are equal to CA, A H respectively. Therefore the triangles FAH, CAH are equal in all respects (Euc. I., 4). Hence FH is equal to HC, and the angles at H are right angles. Thus the triangles D HF and DHC are equal in all respects (Euc. I., 4). Therefore DF is equal to DC. But BD and DF are together greater than BF (Euc. I., 20), that is, than BA, AF together. Therefore BD and DC are together greater than BA and AC together; and the perimeter of BDC is greater than the perimeter of BA C.

Let us next try a few problems-properly so termed that is, propositions in which something is required to be done. In these, as we have said, the analytical method is nearly always to be preferred. We will begin with a simple example.

Ex. 6.- On a given straight line describe an isosceles triangle, each of whose equal sides shall be double of

the base.

Let A B (Fig. 11) be the given straight line.

Suppose that what is required is done, and that on the base AB there has been described the triangle A C B, in which the sides AC and CB are equal to each other, and each double of the base A B; and let us consider what construction is suggested.

It seems hardly possible that the resemblance between this problem and Euc. I., 1, should escape the student's notice. He will inquire, then, whether the method of that problem cannot be applied to the present one. Instead of the circle with radius equal

to A B, we now require circles with radius equal to twice A B. It is clear, then, that if we produce AB to D, making BD equal to AB, and BA to E, making AE equal to AB (Euc. I., 3), then AD and BE will each be double of AB. Therefore if with centre A and radius AD we describe a circle D CH, and with centre B and radius B E the circle ECK, then C, the intersection of these circles, is the vertex of the required triangle. For AC and BC are severally equal to AD and E B—that is, are double of the base, A B.

We will next try the following:

Ex. 7.—The point P (Fig. 12) is within the acute angle formed by the lines A B and A C. It is required to draw through P a straight line which shall cut off equal parts from A B and A C.

Let DPE be the required line, so that AD is equal to A E.2

Then D A E is an isosceles triangle, and it is an obvious course to see whether any of the properties of isosceles triangles will help us to a solution of our problem. Now, the only property of isosceles triangles explicitly contained in Euclid is that of Book I.,

1 We have seen this problem given with the proviso that no problem beyond Euc. I., 1, shall be made use of. In this case the student will see at once that if, with centres A and B, and distance A B, he describes the circles B FE, A G D, then E B and A D, the diameters of these equal circles, are severally double of A B.

2 In constructing the figure, proceed thus:-Take A D equal to A E, and join D E; then take P, a point dividing D E into unequal parts.

Prop. 5. This gives us the angle A D E equal to the angle A E D—a property which avails us nothing.

But there are other properties of isosceles triangles, not expressly mentioned by Euclid, with which every geometrician ought to be acquainted. We will assume that the student is familiar with them--and indeed they are nearly self-evident. They are included in the statement that the perpendicular from the vertex on the base of an isosceles triangle bisects the base and also the vertical angle. Draw

B

P

M

FIG. 12.

A M perpendicular to the assumed line DE; then the angle M AD is equal to the angle M A E, and also DM is equal to ME.

Now let us consider whether this construction

affords us any hints.

First, we cannot see how to draw the line through A perpendicular to the real line D E, because it is this very line we seek to draw.

Secondly, we cannot, for a similar reason, see how to draw the line from A to the bisection of D E.