Cor. 3. If a straight line meet two parallel lines, the opposite exterior and interior angles will be equal. For, the sum HGB+GHD is equal to two right angles. But the sum HGB+EGB is also equal to two right angles. Taking from each the angle HGB, and there remains GHD=EGB. In the same manner we may prove that GHC=AGE.

Scholium. We see that of the eight angles formed by a line cutting two parallel lines obliquely, the four acute angles are equal to each other, and so also are the four obtuse angles.

If two straight lines meet a third line, making the sum of the interior angles on the same side less than two right angles, the two lines will meet if sufficiently produced.

Let the two lines CD, IL, meet the line EF, making the sum of the interior angles HGL, GHD, less than two right angles then will IL and CD meet if sufficiently produced. For, if they do not meet they are parallel (D. 16). But they are not parallel, for if they were, the sum of the interior angles LGH, GHD, would be equal to two right angles (P. 20), whereas it is

less by hypothesis: hence, the

E

I

G

L

-D

C

H

F

lines IL, CD, will meet if sufficiently produced.

Cor. It is evident that the two lines IL, CD, will meet on that side of EF on which the sum of the two angles HGL, GHD, is less than two right angles.

PROPOSITION XXII. THEOREM.

Two straight lines which are parallel to a third line, are parallel to each other.

Let CD and AB be parallel to the third line EF; then are they parallel to each other.

Draw PQR perpendicular to EF, and cutting AB, CD, in the points P and Q. Since AB is parallel to EF, PR will be perpendicular to AB (P. 20, c. 1); and since CD is parallel to EF, PR will for a like reason be

E

R

-D

Q

-B

P

perpendicular to CD. Hence, AB and CD are perpendicular to the same straight line; hence, they are parallel (P. 18).

PROPOSITION XXIII. THEOREM.

Two parallels are everywhere equally distant.

suppose FH and EG These lines will also and we are now to

Let CD and AB be two parallel straight lines. Through any two points of AB, as F and E, to be drawn perpendicular to AB. be perpendicular to CD (P. 20, c. 1); show that they will be equal to each other.

equal to each other (P. 20, c. 2). Also, the straight lines FH, EG, being perpendicular to the same straight line AB, are parallel (P. 18); and the angles EGF, GFH, considered in reference to the parallels FH, EG, will be alternate angles, and therefore equal. Hence, the two triangles EFG, FGH, have a common side, and two adjacent angles in each equal; therefore, the triangles are equal (P. 6); consequently, FH, which measures the distance of the parallels AB and CD at the point F, is equal to EG, which mea sures the distance of the same parallels at the point E

PROPOSITION XXIV. THEOREM.

If two angles have their sides parallel and lying in the same direction, they will be equal.

Let BAC and DEF be the two angles, having AB parallel to ED, and AC to EF; then will they be equal.

For, produce DE, if necessary, till

it meets AC in G. Then, since EF

is parallel to GC, the angle DEF is

equal to DGC (P. 20, c. 3); and since DG is parallel to AB, the angle DGC His equal to BAC; hence, the angle DEF is equal to BAC (A. 1).

D

B

Scholium. The restriction of this proposition to the case where the side EF lies in the same direction with AC, and ED in the same direction with AB, is necessary, because if FE were prolonged towards H, the angle DEH would have its sides parallel to those of the angle BAC, but would not be equal to it. In that case, DEH and BAC would be together equal to two right angles. For, DEH+DEF is equal to two right angles (P. 1); but DEF is equal to BAC: hence, DEH+BAC is equal to two right angles.

PROPOSITION XXV. THEOREM.

In every triangle the sum of the three angles is equal to two

right angles.

Let ABC be any triangle: then will the sum of the angles C+A+B be equal to two right angles. For, prolong the side CA towards

D, and at the point A, suppose AE to be drawn, parallel to BC. Then, since AE, CB, are parallel, and CAD cuts them, the exterior angle DAE is equal to its interior opposite angle

B

E

-D

(P. 20, c. 3). In

like manner, since AE, CB, are parallel, and AB cuts them,

the alternate angles B and BAE, are equal; hence, the three angles of the triangle BAC are equal to the three angles CAB, BAE, EAD, each to each; but the sum of these three angles is equal to two right angles (P. 1); consequently, the sum of the three angles of the triangle, is equal to two right angles (a. 1).

Cor. 1. Two angles of a triangle being given, or merely their sum, the third will be found by subtracting that sum from two right angles.

Cor. 2. If two angles of one triangle are respectively equal to two angles of another, the third angles will also be equal, and the two triangles will be mutually equiangular.

Cor. 3. In any triangle there can be but one right angle: for if there were two, the third angle must be nothing. Still less, can a triangle have more than one obtuse angle.

Cor. 4. In every right-angled triangle, the sum of the two acute angles is equal to one right angle.

Cor. 5. Since every equilateral triangle is also equiangular (P. 11, c. 1), each of its angles will be equal to the third part of two right angles; so, that, if the right angle is expressed by unity, each angle of an equilateral triangle will be expressed by 3.

Cor. 6. In every triangle ABC, the exterior angle BAD is equal to the sum of the two interior opposite angles B and C. For, AE being parallel to BC, the part BAE is equal to the angle B, and the other part DAE is equal to the angle C.

PROPOSITION XXVI. THEOREM.

The sum of all the interior angles of a polygon, is equal to twice as many right angles, less four, as the figure has sides.

Let ABCDE be any polygon: then will the sum of its interior angles

A+B+C+D+E

be equal to twice as many right angles, less four, as the figure has sides.

E

A

D

From the vertex of any angle A, draw diagonals AC, AD, to the vertices of the other angles. It is plain that the polygon will be divided into as many triangles, less two, as it has sides; for, these triangles may be considered as having the point A for a common vertex, and for bases, the several sides of the polygon, excepting the two sides which form the angle A. It is evident, also, that the sum of all the angles in these triangles does not differ from the sum of all the angles in the polygon: hence, the sum of all the angles of the polygon is equal to two right angles, taken as many times as there are triangles in the figure; that is, as many times as there are sides, less two. But this product is equal to twice as many right angles as the figure has sides, less four right angles.

Cor. 1. The sum of the interior angles in a quadrilateral is equal to two right angles multiplied by 4-2, which amounts to four right angles: hence, if all the angles of a quadrilateral are equal, each of them will be a right angle. Hence, each of the angles of a rectangle, and of a square, is a right angle (D. 25).

Cor. 2. The sum of the interior angles of a pentagon is equal to two right angles multiplied by 5-2, which amounts 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 2× (6—2,) or eight right angles; hence, in the equiangular hexagon, each angle is the sixth part of eight right angles, or of one.

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