sum of A D and D C is equal to the sum of AD and D B, which is equal to AB; therefore the side A B is greater than A C.

65. Cor. 1. Therefore the shorter side is opposite to the less angle.

66. Cor. 2. In the right-angled triangle the hypothenuse is the longest side.

PROPOSITION XI.—THEOREM.

67. The greater angle of any triangle is opposite the greater side,

In the triangle CA B, suppose the A side A B to be greater than AC; then will the angle C, opposite to A B, be greater than the angle B, opposite to A C.

с

B

For, if the angle C is not greater than B, it must either be equal to it or less. If the angle C were equal to B, then would the side AB be equal to the side A C (Prop. VIII.), which is contrary to the hypothesis; and if the angle C were less than B, then would the side AB be less than AC (Prop. X. Cor. 1), which is also contrary to the hypothesis. Hence, the angle C must be greater than B.

68. Cor. It follows, therefore, that the less angle is opposite to the shorter side.

69. If, from any point within a triangle, two straight lines are drawn to the extremities of either side, their sum will be less than that of the other two sides of the triangle.

Let the two straight lines BO, CO be drawn from the point O, within the triangle ABC, to the extremities of the side BC; then will the sum of the two lines BO and OC be less than the sum of the sides BA and A C.

B

C

Let the straight line BO be produced till it meets the side A C in the point D; and because one side of a triangle is less than the sum of the other two sides (Prop. IX.), the side OC in the triangle CDO is less than the sum of OD and D C. To each of these inequalities add B O, and we have the sum of BO and O C less than the sum of BO, OD, and DC (Art. 34, Ax. 4); or the sum of BO and O C less than the sum of BD and D C. Again, because the side B D is less than the sum of B A and AD, by adding DC to each, we have the sum of BD and DC less than the sum of BA and A C. But it has been just shown that the sum of BO and OC is less than the sum of BD and DC; much more, then, is the sum of BO and OC less than BA and A C.

70. From a point without a straight line, only one perpendicular can be drawn to that line.

Let A be the point, and DE the given straight line; then from the point A only one perpendicular can be drawn to DE.

Let it be supposed that we can draw two perpendiculars, AB and AC. Produce one of them, as AB, till B F is equal to A B, and join F C.

D

A

E

CB

F

Then, in the triangles A B C and CBF, the angles CBA and CBF are both right angles (Prop. I. Cor. 1), the side CB is common to both, and the side BF is equal to

D

A

E

CB

the side A B; hence the two triangles are equal, and the angle BCF is equal to the angle BCA (Prop. V.) But the angle BCA is, by hypothesis, a right angle; therefore BCF must also be a right angle; and if the two adjacent angles, BCA and BCF, are together equal to two right angles, the two lines AC and CF must form one and the same straight line (Prop. II.). Whence it follows, that between the same two points, A and F, two straight lines can be drawn, which is impossible (Art. 34, Ax. 11); hence no more than one perpendicular can be drawn from the same point to the same straight line.

71. Cor. At the same point C, in the line A B, it is likewise impossible to erect more than one perpendicular to

that line. For, if CD and CE were each perpendicular to A B, the angles A BCD, BCE would be right angles;

F

E D

C

B

hence the angle BCD would be equal to the angle BCE, a part to the whole, which is impossible.

PROPOSITION XIV. -THEOREM.

72. If, from a point without a straight line, a perpendicular be let fall on that line, and oblique lines be drawn to different points in the same line;

1st. The perpendicular will be shorter than any oblique line.

2d. Any two oblique lines, which meet the given line at equal distances from the perpendicular, will be equal. 3d. Of any two oblique lines, that which meets the given line at the greater distance from the perpendicular will be the longer.

they have the side CB common, the side A B equal to the side B F, and the angle A B C equal to the angle F B C, both being right angles (Prop. I. Cor. 1); hence the third sides, CF and A C, are equal (Prop. V. Cor.). But ABF, being a straight line, is shorter than AC F, which is a broken line (Art. 34, Ax. 10); therefore AB, the half of ABF, is shorter than A C, the half of A CF; hence the perpendicular is shorter than any oblique line.

Secondly. If BE is equal to B C, then, since A B is common to the triangles, A BE, ABC, and the angles ABE, ABC are right angles, the two triangles are equal (Prop. V.), and the side A E is equal to the side A C (Prop. V. Cor.). Hence the two oblique lines, meeting the given line at equal distances from the perpendicular, are equal.

Thirdly. The point C being in the triangle A D F, the sum of the lines A C, C F is less than the sum of the sides AD, DF (Prop. XII.) But AC has been shown to be equal to C F; and in like manner it may be shown that A D is equal to DF. Therefore A C, the half of the line A CF, is shorter than AD, the half of the line ADF; hence the oblique line which meets the given line the greater distance from the perpendicular, is the longer.

73. Cor. 1. The perpendicular measures the shortest distance of any point from a straight line.

74. Cor. 2. From the same point to a given straight line only two equal straight lines can be drawn.

75. Cor. 3. Of any two straight lines drawn from a point to a straight line, that which is not shorter than the other will be longer than any straight line that can be drawn between them, from the same point to the same line.

PROPOSITION XV.-THEOREM.

76. If from the middle point of a straight line a perpendicular to this line be drawn,—

1st. Any point in the perpendicular will be equally distant from the extremities of the line.

2d. Any point out of the perpendicular will be unequally distant from those extremities.

Let DC be drawn perpendicular to the straight line AB, from its middle point C.

First. Let D and E be points, taken at pleasure, in the perpendicular, and join DA, DB, and also AE, EB. A4 Then, since AC is equal to CB, the

D

F

E

C

B

two oblique lines D A, D B meet points which are at the same distance from the perpendicular, and are therefore equal (Prop. XIV.). So, likewise, the two oblique lines EA, EB are equal; therefore any point in the perpendicular is equally distant from the extremities A and B.

Secondly. Let F be any point out of the perpendicular, and join F A, FB. Then one of those lines must cut the perpendicular, in some point, as E. Join E B; then we have E B equal to EA. But in the triangle FEB, the side F B is less than the sum of the sides E F, EB (Prop. IX.), and since the sum of FE, EB is equal to the sum of FE, E A, which is equal to FA, FB is less than FA. Hence any point out of the perpendicular is at unequal distances from the extremities A and B.

77. Cor. If a straight line have two points, of which each is equally distant from the extremities of another