Let the straight line DC meet the two straight lines A C, CB at the common point C, making the adjacent angles ACD, DCB together equal to two right angles; then the lines AC and CB will form one and the same straight line.

A

Ꭰ

A

BE

If CB is not the straight line AC produced, let CE be that line produced; then the line ACE being straight, the sum of the angles ACD and DCE will be equal to two right angles (Prop. I.). But by hypothesis the angles ACD and DCB are together equal to two right angles; therefore the sum of the angles A CD and DCE must be equal to the sum of the angles A CD and D C B (Art. 34, Ax. 2). Take away the common angle A CD from each, and there will remain the angle D C B, equal to the angle DCE, a part to the whole, which is impossible; therefore CE is not the line A C produced. Hence AC and CB form one and the same straight line.

PROPOSITION III.-THEOREM.

48. Two straight lines, which have two points common, coincide with each other throughout their whole extent, and form one and the same straight line.

Let the two points which are common to two straight lines be A and B.

F

E

A

D

B C

The two lines must coincide between the points A and B, for otherwise there would be two straight lines between A and B, which is impossible (Art. 34, Ax. 11).

Suppose, however, that, on being produced, the lines begin to separate at the point C, the one taking the direc

tion CD, and the other C E. From the point C let the line CF be drawn, making, with CA, the right angle ACF. Now, since ACD is a straight line, the angle FCD will be a right angle (Prop. I. Cor. 1); and since ACE is a straight line, the angle FCE will also be a right angle; therefore the angle FCE is equal to the angle FCD (Art. 34, Ax. 13), a part to the whole, which is impossible; hence two straight lines which have two points common, A and B, cannot separate from each other when produced; hence they must form one and the same straight line.

PROPOSITION IV. THEOREM.

C

49. When two straight lines intersect each other, the opposite or vertical angles which they form are equal. Let the two straight lines AB, CD intersect each other at the point E; then will the angle AEC be equal to the angle DEB, and the angle CEB to AED.

E

B

D

For the angles AEC, CEB, which the straight line C E forms by meeting the straight line AB, are together equal to two right angles (Prop. I.); and the angles CEB, BED, which the straight line BE forms by meeting the straight line CD, are equal to two right angles; hence the sum of the angles A EC, CEB is equal to the sum of the angles CEB, BED (Art. 34, Ax. 1). Take away from each of these sums the common angle CE B, and there will remain the angle A E C, equal to its opposite angle, B E D (Art. 34, Ax. 3).

In the same manner it may be shown that the angle CEB is equal to its opposite angle, A ED.

50. Cor. 1. The four angles formed by two straight lines intersecting each other, are together equal to four right angles.

*

51. Cor. 2. All the successive angles, around a common point, formed by any number of straight lines meeting at that point, are together equal to four right angles.

PROPOSITION V.-THEOREM.

A

D

52. If two triangles have two sides and the included angle in the one equal to two sides and the included angle in the other, each to each, the two triangles will be equal. In the two triangles ABC, DEF, let the side A B be equal to the side DE, the side A C to the side D F, and the angle A to the angle D;

Дл

B

C E

then the triangles ABC, DEF will be equal.

Conceive the triangle ABC to be applied to the triangle DEF, so that the side A B shall fall upon its equal, D E, the point A upon D, and the point B upon E; then, since the angle A is equal to the angle D, the side AC will take the direction DF. But AC is equal to DF; therefore the point C will fall upon F, and the third side BC will coincide with the third side E F (Art. 34, Ax. 11). Hence the triangle A B C coincides with the triangle DEF, and they are therefore equal (Art. 34, Ax. 14).

53. Cor. When, in two triangles, these three parts are equal, namely, the side AB equal to DE, the side A C equal to D F, and the angle A equal to D, the other three corresponding parts are also equal, namely, the side BC equal to EF, the angle B equal to E, and the angle C equal to F.

PROPOSITION VI.—THEOREM.

54. If two triangles have two angles and the included side in the one equal to two angles and the included side in the other, each to each, the two triangles will be equal.

then the triangles A B C, DEF will be equal.

Conceive the triangle A B C to be applied to the triangle DEF, so that the side B C shall fall upon its equal, EF, the point B upon E, and the point C upon F. Then, since the angle B is equal to the angle E, the side BA will take the direction ED; therefore the point A will be found somewhere in the line ED. In like manner, since the angle C is equal to the angle F, the line CA will take the direction FD, and the point A will be found somewhere in the line F D. Hence the point A, falling at the same time in both of the straight lines ED and FD, must fall at their intersection, D. Hence the two triangles ABC, DEF coincide with each other, and are therefore equal (Art. 34, Ax. 14).

55. Cor. When, in two triangles, these three parts are equal, namely, the angle B equal to the angle E, the angle C equal to the angle F, and the side B C equal to the side EF, the other three corresponding parts are also equal; namely, the side B A equal to ED, the side CA equal to FD, and the angle A equal to the angle D.

PROPOSITION VII.-THEOREM.

56. In an isosceles triangle, the angles opposite the equal sides are equal.

Let ABC be an isosceles triangle, in which the side AB is equal to the side AC; then will the angle B be equal to the angle C.

Conceive the angle B A C to be bisected, or divided into two equal parts, by

А

D

C

the straight line AD, making the angle
BAD equal to DAC. Then the two
triangles BAD, CAD have the two
sides A B, AD and the included angle
in the one equal to the two sides A C,
AD and the included angle in the other, B

each to each; hence the two triangles are equal, and the angle B is equal to the angle C (Prop. V.).

57. Cor. 1. The line bisecting the vertical angle of an isosceles triangle bisects the base at right angles.

58. Cor. 2. Conversely, the line bisecting the base of an isosceles triangle at right angles, bisects also the vertical angle.

59. Cor. 3. Every equilateral triangle is also equiangular.

PROPOSITION VIII.-THEOREM.

A

60. If two angles of a triangle are equal, the opposite sides are also equal, and the triangle is isosceles. Let A B C be a triangle having the angle B equal to the angle C; then will the side A B be equal to the side A C.

B

D

C

For, if the two sides are not equal, one of them must be greater than the other. Let A B be the greater; then take D B equal to AC the less, and draw CD. Now, in the two triangles DB C, ABC, we have D B equal to AC by construction, the side BC common, and the angle B equal to the angle ACB by hypothesis; therefore, since two sides and the included angle in the one are equal to two sides and the included angle in the other, each to each, the triangle DBC is equal to the triangle A B C (Prop. V.), a part to the whole, which is impossible (Art. 34, Ax. 8). Hence the sides A B and AC cannot be unequal; therefore the triangle ABC is isosceles.