=

line of 6 feet is 36; the square of the part AB = 2 is 4; the square of BC 4 is 16; the rectangle, or product of AB=2 by BC=4, is 8, and twice this product or rectangle is 16: add, therefore, these two squares and the two products together; 4+16+8+8=36, which is equal to the square of the whole line AB, or 6×6=36.

PROP. XVIII. fig. 1, Plate 2. The square constructed on the hypothenuse of a right-angled triangle is equivalent to the sum of the squares constructed or the two sides containing the right angle. Let ABC be a triangle, having a right angle at B; on the side AB construct the square AFGB, on BC the square BHIC, and on the hypothenuse AC the square ACDE; from the right angle at B, let fall on the hypothenuse the perpendicular BK, and produce it to L, and join BE and FC. The angle BAE is composed of the angles BAC and CAE, of which CAE being the ang'e of a square, must be a right angle: again, the angle FAC is composed of FAB and BAC, of which FAB, the angle of a square, must be a right angle, therefore equal to CAE, and BAC is common to both; consequently the whole angle FAC will be equal to the whole angle BAE. Then AC and AE being sides of a square, they are equal; and for the same reason AB is equal to AF; we have therefore two triangles, CFA and ABE, having two sides of the one equal to two sides of the other, and the contained angle of the one FAC equal to the contained angle of the other BAE; these two triangles must consequently be equal (Prop. 3, fig. 17, Plate 1). But the triangle CFA is one half of the square AFGB, for it stands on the same base, and within the same parallels, AF and GBC, (Prop. 12 and 15, fig. 27); and for the same reason the triangle ABE is one half of the parallelogramı AKLE; but if the halves of any magnitudes be equal, the wholes must also be equal; therefore the whole square AFGB must be equal to the whole parallelogram AKLE.

Again, by drawing the diagonal lines AI and BD, it may in the same way be demonstrated, that the triangle AIC is equal to the triangle BDC; and, consequently, the square BHIC to the parallelogram KCDL : but it was already shown, that AKLE is equal to the square AFGB, and now it appears that KCDL is equal to the square BHIC: and as these two parallelograms, AKI.E and KCDL, compose the square ACDE, it follows, that the square ACDE, is equal to the sum of the two squares AFGB and EHIC : or, in other words, that the square of the hypothenuse of a rightangled triangle is equal to the sum of the squares of the two sides containing the right angle.

To illustrate this proposition by arithmetic, let the side AB be equal to 3, the side BC to 4, and the hypothenuse AC to 5; the square of 3 is 9; the square of 4 is 16; and the sum of these squares is 25, equal to the square of the hypothenuse 5.

N. B. These three numbers, 3, 4, and 5, or any multiples of them, will always form a right-angled triangle: thus, if the sides were 15, 20, and 25, which are 5 times 3, 4, and 5, the square of 15 which is 225, added to the square of 20, or 400, will give 625, which is just the square of the hypothenuse 25.

Corollary. The square of a side opposite to an angle greater than a right angle, will be greater than the sum of the squares of the two sides containing the obtuse angle: and if the angle be acute, or less than a right angle, the square of the opposite side will be less than the sum of the squares of the containing sides.

PROP. XIX. fig. 32. If in a triangle, as ABC, a right line be drawn, as DE, parallel to one of the sides, as AC, and cutting the two other sides in D and E, this line will cut these sides proportionally, that is, BD will be to BE as BA is to BC. Join AE and DC, and the triangles AED

and CDE being constructed on the same base DE, and of the same altitude (Prop. 12, fig. 27), they are equal: and the triangles BDE, CDE, being of the same altitudes, are to each other as their bases; that is, the triangle BDE is to the triangle CDE, as BE to CE: but CDE has just been shown to be equal to AED, and AED is to BED as the base AD to the base BD: hence we have the four proportionals BE: EC:: BD: DA: and by compounding the proportions BE: BC:: BD: DA, or BE: BD :: BC: BA.

The reverse of this proposition is also true: that is, if two sides of a triangle are cut, proportionally, by a right line drawn within it, such a line must be parallel to the third side of the triangle.

PROP. XX. fig. 33, Plate 1. If a right line BD be drawn in a triangle ABC bisecting, that is, dividing into two equal parts any angle, as that at B, and cutting the opposite side AC, in the point D, then will this side AC be cut into two parts proportionally to the opposite sides of the triangle forming the bisected angle; that is, the part of the base AD will be to the part DC, as the side AB to the side BC.

From A draw AE parallel to BD, and meeting CB produced in the point E; and, by Prop. 19, we shall have CD: DA:: CB: BE: but from the properties of parallel lines formerly explained (Prop. 6, fig. 20), the angle DBA will be equal to BAE, and DBA will also be equal to BEA: BAE and BEA are therefore equal to each other, and consequently the opposite sides BE and BA must be equal; and hence CB: BE:: CB: BA, and CD: DA:: CB: BA; that is, the segments of the base of the triangle, made by a line bisecting the angle opposite to the base, will be proportional to the respective sides forming that angle.

PROP.

PROP. XXI. fig. 34. If two triangles, as ABC and DEF, have any angle of the one equal to an angle of the other, as the angles at B and E, and the sides forming these angles respectively proportional, that is, AB: BC:: DE: EF; then these triangles will be similar to each other. In the greater triangle DEF, make EH equal to BC of the less triangle, and draw HG parallel to the side DF, then will the angle EHG be equal to the angle EFD, and the triangle EGH be equiangular to the triangle EFD; and consequently EF: EH:; ED: EG: but by the proposition EF: BC:: ED; BA, and EH being made equal to BC, EG will be equal to BA, and the triangles EHG and BCA will be equal: but the triangle EHG was already shown to be similar to EFD; therefore the triangle ABC will be similar to the triangle DEF.

From these properties of similar triangles it may be demonstrated, that the areas of such triangles are to each other, in the proportion of the squares of their corresponding or homologous sides; that is, (fig. 34) the area of the triangle ABC, will be to that of the triangle DEF, as the square of AB is to the square of DE, or as the square of BC to the square of EF.

To bisect a given right

PROP. XXII. fig. 2, Plate 2. To bisect a line, AB, that is, to cut it into two equal parts. Upon the extremity A, as a centre, with a radius or opening of the compasses greater than half of AB, make arches on each side of that line, as a b, cd; from B, as a centre, with the same radius, describe the arches ef, gh, intersecting the former arches in the points I and K: then lay a ruler, or draw a line from I to K, cutting AB in the point O; so will AB be bisected at O, that is, AO will be equal to OB. For the points I and K being found by intersecting arches described with the same radius, these points must be equally distant from the extremities A and B ; and

a right

a right line joining I and K, will also, in every other point, be equally distant from A and B; consequently the point O, where IK cuts AB, must be equally distant from A and B, that is, the line AB must be cut into two equal parts at the point O.

PROP. XXIII. fig. 3. From a given point, C, in the right line, AB, to draw a right line, CF, which shall be perpendicular to AB. Take the points D and E in AB, equally distant from the point C; then from D and E, with any radius greater than DC or CE, make the intersection F, and join FC: then will FC be perpendicular to AB, and it is drawn at the given point C, which was the thing required to be done. See Prop. 22.

PROP. XXIV. fig. 4. From a given point, C, to let fall a perpendicular on a given line, AB. From C, as a centre, with a radius reaching beyond AB, describe an arch cutting AB in two points; and from these points, with any convenient radius, make the intersection D on the side opposite to C: a line drawn from C towards D, till it touch AB, in the point E, will be perpendicular to AB; and it is drawn from the point C, as was required to be done.

PROP. XXV. fig. 5. To erect a perpendicular on the extremity B, of a given right line, AB. Choose any point, C, such that a perpendicular let fall from it, would come within the point B: from C, with the radius CB, describe the arch DBE, cutting AB in D; draw a line joining D and C, and produced till it cut the arch in E; then the segment, or arch, DBE, being a semicircle, the angle at B is a right angle, consequently the line EB is drawn perpendicular to the line AB, at the point B, as was required.

PROP.