PROPOSITION 1. THEOREM.

In any right-angled triangle, the square which is described on the side subtending the right angle is equal to the sum of the squares described on the side which contains the right angle.

Let ABC be a right-angled triangle, having the right angle BAC; the square described on the side BC shall be equal to the sum of the squares described on BA, AC. See Plate 4 Fig. 1. On BC describe the square

squares GB, HC;

BDEC, and on BA, AC describe the

through A draw AL parallel to BD or CE; and join AD, EC.

Then, because the angle BAC is a right angle, and that the angle BAG is also a right angle,

Todhunter's Euclid, [1.46,]

[1.31,]

[Hypothesis,]

[Definition 30,]

the two straight lines AC, AG, on the opposite sides of AB, make with it at the point A the adjacent angles equal to two right angles; therefore CA is in the same straight line with AG.

[1.14.] For the same reason, AB and AH are in the same straight line. Now the angle DBC is equal to the angle FBA, for each of them is

a right angle.

[Axiom 11.] Add to each the angle ABC. Therefore the whole angle DBA is equal to the whole angle FBC.

[Axiom 2.]

And because the two sides AB, BD are equal to the two sides FB, BC, each to each;

[Definition 30,]

and the angle DBA is equal to the angle FBC; therefore the triangle ABD is equal to the triangle FBC.

[1.4.]

Now the parallelogram BL is double of the triangle ABD, because they are on the same base BD, and between the same parallels BD, AL.

[1.41.] And the square GB is double of the triangle FBC, because they are on the same base FB, and between the same parallels FB, GC, [1.41.] But the doubles of equals are equal to one another; [Axiom 6.] Therefore the parallelogram BL is equal to the square GB.

In the same manner, by joining AE, BK, it can be shown that the parallelogram CL is equal to the square CH.

Therefore the whole square BDEC is equal to the two squares GB, HC.

[Axiom 2.]

And the square BDEC is described on BC, and the squares GB, HC

on BA, AC. Therefore the square described on the side BC is equal to the squares described on the sides BA, AC.

CORALLARY 1. Let ABCD be a square, and AC its diagonal; the triangle ABC being right-angled and isoceles, we have AC2=AB2+BC2 -2AB2.

Therefore, the square described on the diagonal of a square, is double of the square described on the side.

If we extract the square root of each member of this equation, we shall have

ARGUMENT 1. If ten (10) right-angled triangles, the base, perpendicular, and hypothenuse of which are respectively three (3), four (4), and five (5), be placed so as to form a polygon of five sides and a little more, that is to say, with all their bases outwards, and the perpendicular of one triangle joined to the perpendicular of another, so that they shall touch one another in those points which contain the right angles, and also in those points which form the verticle angles, so that any two of the bases when taken together shall be equal to six; and the hypothenuse of each triangle joined to the hypothenuse of another triangle, so that the vertical angles, that is the angles opposite the bases shall form a common center, then there will be a lap equal to the one-fiftieth of the circumference of the whole polygon. See Plate 5 Fig. 1. For, because the straight lines AC, CD, are tangents to the radius BC, therefore they are both in the same straight line, and the radius BC is a perpendicular to them. For the same reason the straight lines DE, EF are in the same straight line, and the radius BE is a perpendicular to them, also FG, GH, and BG. HI, IJ, and BI. JK, KZ. and BK. [Hypothesis.] Again, because the radius BC is perpendicular to the straight line AD, therefore, the angles BCD and BCA are both right angles; for the same reason BED and BEF are right angles, as also BGF and BGH, BIH, and BIJ, BKJ and BKZ. [Hypothesis.]

Again, because the angles ADF, DFH, FHJ, HJZ are inscribed in an arc which is less than a semicircle, therefore they are greater than a right angle.

[Hypothesis.]