PROPOSITION G. THEOREM.

ED.

A straight line passing through the middle of the diagonal of a parallelogram divides the parallelogram into two equal parts.

Let ABDC be a paral., AD a diagonal, and EF a line passing through its middle G; the figure ABFE is equal to EFDC. For the triangles ABD, ACD are equal, and the F triangles AEG, DGF are equiangular, and have the sides AG, DG equal, therefore they are equal (Prop. A). Conseq. if they be

taken from the triangles ABD, ACD, the remainders ABFG, CDGF will be equal. Hence it is evident that ABFG and AGE are together equal to CDGE and DGF together, or ABFE, CDFE. Therefore, a straight line &c. Q. E. D.

Cor. Any straight line EF drawn through the middle of the diagonal of a parallelogram is bisected by the diagonal.

PROPOSITION H. THEOREM.

ED.

If two sides of a triangle be bisected, the straight line which joins the points of section will be parallel to the third side, and equal to half of it.

Let the sides AB, AC of the triangle ABC be bisected in the points D, E; the straight line DE joining those points will be parallel to BC, and equal to half of BC. In DE produced take EF equal to DE, B and join CF. Because

the side DE is equal to EF, and AE equal to EC, and the angle AED equal to CEF, the triangles ADE, FEC are equal in all respects, therefore the side AĎ is equal to CF, and the angle ADF equal to the alternate angle CFE, therefore AB is parallel to FC.

Because BD or AD and CF are equal and parallel, the straight lines BC, DF joining their extremes are also equal and parallel (Prop. 33). But DF is double of DE, therefore, BC is double of DE. Therefore, if two sides &c. Q. E. D.

PROPOSITION I. THEOREM.

ED.

The diagonals of any parallelogram bisect each other, and divide the parallelogram into four equal triangles.

1. Let ABCD be any parallelogram, of which the diagonals are AC and BD; they will bisect each other.

Let AC, BD intersect each other in E; then, in the triangles ADE, BCE, the vertical angles at E are equal (Prop. 15), and the alternate angles EAD, ECB are equal (Prop. 29), and the alter- B nate angles EDA, EBC

A

F

C

are equal, and the sides AD, BC are equal (Prop. 34); therefore the other sides which are opposite to equal angles are also equal (Prop. A), namely, AE to EC, and DE to EB. Therefore the diagonals AC, BD bisect each other in E.

2. The diagonals AC, BD divide the parallelogram into four equal triangles.

For the triangles ABE, ADE being on equal bases BE, ED (by case 1), and between the same parallels, are equal; and the triangles ADE, CDE being on equal bases AE, EC, and between the same parallels, are equal; and the triangles CDE, CBE being on equal bases DE, EB, and between the same parallels, are equal. Consequently the four triangles ABE, ADE, CDE, CBE, are equal to one another. Therefore, the diagonals &c. Q. E. D.

PROPOSITION K. THEOREM.

ED.

The diagonals of a square bisect each other, and divide the square into four right angled triangles which are equal in all respects.

Let ABCD be a square, of which the diagonals are AC, BD; the four triangles ABE, ADE, CBE, CDE, are right angled, and are equal in all respects.

For the diagonals AC, BD bisect each other (Prop. I.), and divide the square into four triangles which are mutually equilateral, and therefore are equal in all respects, therefore the angles at E are right angled. Therefore, the diagonals &c. Q. E. D.

A

E

D

B

C

PROPOSITION L. THEOREM.

All the interior angles of any rectilineal figure are together equal to twice as many right angles as the figure has sides, wanting four right angles.

Let ABCDE be any rectilineal figure; all its interior angles A, B, C, D, E, are together equal to twice as many right angles, wanting four, as the figure has sides.

For any rectilineal figure ABCDE can be divided into as many triangles as it has sides, by drawing lines from a point F within the figure to each of its angles; and the three angles of each triangle are together equal to two right angles (Prop. 32), therefore all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure. But the same angles are equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles, that is, together with four right angles (3 Cor. 13). Therefore, twice as many right angles as the figure has sides are equal to all the angles of the figure, together with four right angles: that is, the angles of the figure are equal to twice as many right angles as the figure has sides, wanting four. Therefore, all the interior angles &c. Q. E. D.

COR. All the interior angles of any quadrilateral figure are together equal to four right angles.

Scholium. Let n denote the number of the sides of a polygon, s the sum of all the interior angles, and r a right angle; then the proposition will be expressed by this theorem, s=2r xn-4r, by means of which the sum of all the interior angles of any convex polygon may be found. Thus, if the figure be quadrilateral, then n=4, and s = 2r x n − 4r 4r, that is, four right angles, which answers to 6 Cor. 32. If n 5, then 2r x n— - 4r= 10r4r6r.

-

= 8r 4r

If the figure proposed be an equiangular polygon, or have all its angles equal, the quantity of each angle may be found by this formula. Suppose an equiangular polygon of six sides, then 2r x n — 4r : 12r4r= Sr, therefore r = r = x 90 4 x 30 = 120, that is, each of the angles of a regular hexagon is 120 degrees. Let the figure be a regular octagon, then 2r X N· 4r=12r, therefore 12r3r3 × 90 = 3 x 45 135 degrees.

=

Lets=2rxn-4r=(2n−4) xr=(n-2) × 2r, that is, all the interior angles of any convex rectilineal figure are together equal to twice as many right angles as the figure has sides less two sides. ED.

PROPOSITION M. THEOREM.

All the exterior angles of any convex rectilineal figure, made by producing all the sides outward in the same direction, are together equal to four right angles.

Let A, B, C, &c. be the exterior angles of any rectilineal figure, made by producing its sides; the exterior angles A, B, C, &c. are together equal to four right angles.

Because every interior angle ABC of the figure, with its adjacent exterior angle ABD, are equal to two right angles (Prop. 13); and because there are as many exterior or interior angles as the figure has sides; all the interior together with all the exterior angles of the figure are equal to twice D as many right angles as there are sides of the figure. But all the interior angles together with four right angles

B

are equal to twice as many right angles as there are sides of the figure (Prop. L). Therefore all the interior and all the exterior angles are together equal to all the interior angles together with four right angles. From these equals take away all the interior angles, and all the exterior angles will be equal to four right angles. Therefore, all the exterior &c. Q. E. D.

Scholium. The propositions in this Book are not arranged in the order of the subjects, but in such order as to enable the Author to demonstrate certain propositions by means of others which are placed before them. In the following abstract they are disposed according to the nature of the subjects, and such theorems as are of little or no use are omitted. ED.

THE PRINCIPAL THEOREMS IN BOOK I.

Properties of Straight Lines and Angles:

If one straight line meet another, the sum of the two adjacent angles is equal to two right angles.

If any number of straight lines intersect or meet one another in the same point, the sum of all the angles about that point is equal to four right angles.

If two straight lines intersect each other, the vertical angles are equal.

Properties of Parallel Lines.

If a straight line intersect two straight lines, and make the alternate angles equal, or the exterior angle equal to the interior and opposite angle on the same side of the cutting line, or the sum of the two interior angles on the same side equal to two right angles, then those two lines are parallel.

A straight line intersecting two parallel straight lines makes the alternate angles equal, and the exterior angle equal to the interior and opposite angle on the same side of the cutting line, and the sum of the two interior angles on the same side equal to two right angles. Straight lines which are parallel to the same straight line are parallel to one another.