Cor. 1. The sum of the interior angles of a quadrilateral is equal to two right angles taken twice; that is, to four right angles. If the angles of a quadrilateral are equal, each will be a right angle.

Cor. 2. The sum of the interior angles of a pentagon is equal to two right angles taken three times; that is, to six right angles: hence, when a pentagon is equiangular, each angle is equal to the fifth part of six right angles, or to g of one right angle.

Cor. 3. The sum of the interior angles of a hexagon is equal to eight right angles : hence, in the equiangular hexagon, each angle is the sixth part of eight right angles, or of one right angle.

Cor. 4. In any equiangular polygon, any interior angle is equal to twice as many right angles as the figure has sides, less four, divided by the number of angles.

The sum of the exterior angles of a polygon is equal to four right angles.

Let the sides of the polygon ABCDE be prolonged, in the same order, forming the exterior angles a, b, c, d, e; then will the sum of these exterior angles be equal to four right angles.

For, each interior angle, together with the corresponding exterior angle, is equal

to two right angles (P. I.): hence, the sum of all the inte rior and exterior angles is equal to two right angles taken

as many times as the polygon has sides.

But the sum of

the interior angles is equal to two right angles taken as many times as the polygon has sides, less two: hence, the sum of the exterior angles is equal to two right angles taken twice; that is, equal to four right angles; which was to be proved.

PROPOSITION XXVIII. THEOREM.

In any parallelogram, the opposite sides are equal, each to

each..

Let ABCD be a parallelogram: then will AB be equal to DC, and AD to BC.

For, draw the diagonal BD. Then, because AB and DC are parallel, the A angle DBA is equal to its alternate angle BDC (P. XX., C. 2) and, because AD and BC are parallel, the angle BDA is equal to its alternate angle DBC. The triangles ABD and CDB, have, therefore, the angle DBA equal to CDB, the angle BDA equal to DBC, and the included side DB common; consequently, they are equal in all of their parts: hence, AB is equal to DC, and AD to BC; which was to be proved.

P

Cor. 1. A diagonal of a parallelogram divides it into two equal triangles.

Cor. 2. Two parallels included between two other par llels, are equal.

Cor. 3. If two parallelograms have two sides and the included angle of the one, equal to two sides and the ineluded angle of the other, each to each, they will be equal.

If the opposite sides of a quadrilateral are equal, each to each, the figure is a parallelogram.

In the quadrilateral ABCD, let AB be equal to DC, and AD to BC: then will it be a parallelogram.

Draw the diagonal DB. Then, the triangles ADB and CBD, will have

A

D

the sides of the one equal to the sides of the other, each to each; and therefore, the triangles will be equal in all of their parts hence, the angle ABD is equal to the angle CDB (P. X., S.); and consequently, AB is parallel to DC (P. XIX., C. 1). The angle DBC is also equal to the angle BDA, and consequently, BC is parallel to AD: hence, the opposite sides are parallel, two and two; that is, the figure is a parallelogram (D. 28); which was to be proved.

PROPOSITION XXX. THEOREM.

If two sides of a quadrilateral are equal and parallel, the figure is a parallelogram.

In the quadrilateral ABCD, let AB

D

be equal and parallel to DC: then will

the figure be a parallelogram.

Draw the diagonal DB. Then, because AB and DC are parallel, the

A

angle ABD is equal to its alternate angle CDB. Now, the triangles ABD and CDB, have the side DC equal to AB, by hypothesis, the side DB common, and the included angle ABD equal to BDC, from what has just

been shown; hence, the triangles are equal in all their parts (P. V.); and consequently, the alternate angles ADB and DBC are equal. The sides BC and AD are, therefore, parallel, and the figure is a parallelogram; which was to be proved.

Cor. If two points be taken at equal distances from a line, and on the same side of it, the line joining them will be parallel to the given line.

PROPOSITION XXXI. THEOREM.

The diagonals of a parallelogram divide each other into equal parts, or mutually bisect each other.

Let ABCD be a parallelogram, and AC, BD, its diagonals: then will AE be equal to EC, and BE BE to ED. For, the triangles BEC and AED, have the angles EBC and ADE equal

(P. XX., C. 2), the angles ECB and DAE equal, and the included sides BC and AD equal: hence, the triangles are equal in all of their parts (P. VI.); consequently, AE is equal to EC, and BE to ED; which was to be proved

Scholium. In a rhombus, the sides AB, BC, being equal, the triangles AEB, EBC, have the sides of the one equal to the corresponding sides of the other; they are, therefore, equal: hence, the angles AEB, BEC, are equal, and therefore, the two diagonals bisect each other at right angles.

BOOK II.

RATIOS AND PROPORTIONS.

DEFINITIONS.

1. THE RATIO of one quantity to another of the same kind, is the quotient obtained by dividing the second by the first. The first quantity is called the ANTECEDENT, and the second, the CONSEQUENT.

A PROPORTION is an expression of equality between two equal ratios. Thus,

expresses the fact that the ratio of A to B is equal to the ratio of C to D. In Geometry, the proportion is written thus,

and read, A is to B, as C is to D.

3. A CONTINUED PROPORTION is one in which severa ratios are successively equal to each other; as,

A: B : CD: E

F:: G

H, &c.

4. There are four terms in every proportion.

The first

and second form the first couplet, and the third and fourth,