Fig. 69.

Because, in the triangle LOC, the side LO is to LC, and the angle OLC a right angle, the angle LCO is half a right angle; for the same reason the angle FCD is half a right angle, and the angle LCF is a right angle; ... the three LCO, LCF, and FCD are = to two right angles; .. (prop. 14, 1 Elr.) the right lines OC and CD are in directum.

PROP. 151. THEOR.

In figure sixty-nine, the diagonal (CV) of the parallelogram (LCFV), formed by producing the sides (DF and LV) of the squares described on the sides of a right angled triangle, is in directum with the perpendicular (CT) let fall from the vertical angle on the hypotenuse. Because, in the triangles CLV, and BCA, the side CL is to BC, the side LV (= CF) = CA, and the angle CLV BCA, the angle ABC shall be to VCL; but the angle ABC with BCT is to a right angle; .. VCL and BCT are to a right angle; but LCB is a right angle; .. the three TCB, BCL, and LCV are right angles; .. TC and CV are in directum.

PROP. 152. THEOR.

to two

If right lines (DP, KG, and LI), drawn at right angles to the sides of an equilateral triangle (ABC), form a triangle (POF), it also shall be equilateral.

Fig. 74.

=

Because ILC is a right angle, it is to the sum of the angles LCI and CIL; but LCI is an angle of 60 degrees; ... CIL is an angle of 30°; and since ODI is a right angle, and DIO an angle of 30°, IOD is an angle of 60°; .. its vertically opposite angle POF is an angle of 60°; for the same reason OFP is an angle of 60°; .. POF is an equilateral triangle.

PROP. 153. THEOR.

If right lines (RT, IH, and US) be drawn from the sides of any triangle (OPQ) with the same inclination, i. e. making the angles (TRP, HIQ, and SUO) equal to one another, the triangle (e d o) formed by them shall be equiangular to the whole triangle.

Fig. 75.

For the angle TRP is to the angles RTQ + RQT; but HIQ is to TRP, .. HIQ is = to RTQ + RQT; and also to I To+IOT;.. IOT is to RQT;.. its vertically opposite angle doe is to PQO; in like manner e do can be proved to OPQ; .. o ed is to POQ; ... &c. &c.

PROP. 154. THEOR.

In figure sixty-eight, the parallelogram (ABCD) is equal to half the difference of the triangles (EGF and HGI.)

For, draw EH and FI and bisect them in 6 and d, join b, B and d, C and also A b, b D, D d, and d A.

Then it is evident, that the triangles A C d, and Bb D are ; ... they are together to half the triangle IEF; for the same reason the triangles CDd, and 6 B A are together to half the triangle HEF; and the parallelogram Ab D d is to half the figure EHIF. Then EHIF is to HEG + IGF + EGF + HGI, which are = to twice Ab Dd, which is to HEG + EGF + IGF + EGF twice ABDC, since the four small triangles D Bb, b B A, Ad C, dC D are together to half the sum of the triangles HEF, IEF; .. HEF + IEF + twice ABDC is to twice Ab D d or to EHIF; but IGH + HGE + EGF + FGI is also to EHIF; IGH and EGF are

.. HGI

EGF is

to twice EGF + twice ABDC to twice ABDC;.'. ABDC is = to half the difference of the triangles HGI, and EGF.

END OF FIRST BOOK.

THE

SECOND BOOK,

IN

GENERAL TERMS.

PROPOSITION 1. THEOR.

If there be two right lines, one of which is divided into any number of parts, the rectangle under the two lines is equal to the sum of the rectangles under the undivided line and the several parts of the divided line.

Draw from either extremity of the divided line, a line at right angles to it, and to the undivided line, and complete the rectangle; then, through each point of bisection of the divided line, draw a line parallel to the perpendicular, to meet the opposite side.

Then it is evident, that those parallel lines are = to the perpendicular, and.. to the undivided line; also that the whole rectangle is to a rect. under the given lines, and is divided into as many rectangles as there are segments in the divided line, and.. into rectangles under the undivided line and the several parts of the divided line.

Cor. Hence and from prop. 34, b. 1, it appears, that the area of a rectangle is found by multiplying its altitude into its base; and from prop. 35 and 36, b. I, it also appears, that the area of any parallelogram is found by multi

N

plying its altitude into its base; § and from prop. 37, b. 1, that the area of a triangle is found by multiplying its altitude into half its base.

Those propositions may be reduced to numbers, thus: suppose the divided line 16 feet is divided into 8, 6, and 2 feet; then 16 multiplied by the undivided line 6 feet, is to the sum of the products of 8 x 6, 6 x 6 and 2 x 6.

=

For 16 x 6 96, and 8 x 6 48, +6 x 6 = 84, + 2 × 6 = 96.

From this it is evident, that the product of any two numbers is to the sum of the products of either of them multiplied into the several parts, into which the other may be divided; for, if 246 is to be multiplied by 6, the product will be the same as the sum of the products of 200 x 6, 40 x 6 and 6 x 6; from this principle, the arithmetical process of multiplication is performed, which is nothing more than common addition; for 246 multiplied by 6 is the same as 246 placed 6 times under one another and the entire added together.

† For if the altitude be 8 feet and the base 10, it will be to 10 rectangles, each of whose bases is one foot and altitude eight; the area of each of these would be 8 square feet; .. the area of the whole rect. will be 80 square feet. From this it is evident, that the rect, under any line and its tenth part is to ten times the of that part.

For it is to the area of a rectangle on the same base and of the same altitude, i. e. between the same parallels.

For it is to half the area of a parallelogram on the same base and of the same altitude.

PROP. 2. THEOR.

If a right line be divided into any two parts, the square of the whole line is equal to the sum of the rectangles under the whole line and each of the parts.

2

Describe on the given line a (prop. 46, 1.), and through the point of section draw a line parallel to one of the sides.

Then it is evident (prop. 1, b. 2.), that this is = to