+DC<BA+AC. But it has just been found that BO+OC <BD+DC; therefore, still more is BO+OC<BA+AC.

PROPOSITION XII. THEOREM.

If two sides of one triangle be respectively equal to two sides of another, but include a greater angle; the third side of the former shall exceed the third side of the latter.

Let BAC and EDF be two triangles, having the side AB -DE, AC-DF, and the angle AD; then will BC> EF.

Make the angle CAG=D; take AG-DE, and draw

CG. The triangle GAC is equal to DEF, since, by construction, they have an equal angle in each, contained by equal sides, (Prop. VIII.); therefore CG is equal to EF. Now, there may be three cases in the proposition, according as the point G falls without the triangle ABC, or upon its base BC, or within it.

First Case. The straight line GC GI+IC, and the straight line AB<AI+IB; therefore, GC+AB<GI+AI+ IC+IB, or, which is the same thing, GC+AB<AG+BC. Take away AB from the one side, and its equal AG from the other; and there remains GC<BC (Ax. 5.); but we have found GC-EF, therefore BCEF.

Second Case. If the point G fall on the side BC, it is evident that GC, or its equal EF, will be shorter than BC, since BC=BG+GC.

Third Case. Lastly, if the point G fall within the triangle BAC, we shall have, by the preceding theorem, AG+GC<AB+ BC; and taking AG from the one, and its equal AB from the other, there will remain GC BC, or BC EF.

Scholium. Conversely, if two sides BA, AC, of the triangle BAC, are equal to the two ED, DF, of the triangle EDF, each to each, while the third side BC of the first triangle is greater than the third side EF of the second; then will the angle BAC of the first triangle be greater than the angle EDF of the second.

B

For, if not, the angle BAC must be equal to EDF, or less than it. In the first case, the side BC would be equal to EF, (Prop. V.); in the second, CB would be less than EF; but either of these results contradicts the hypothesis; therefore, BAC is greater than EDF.

PROPOSITION XIII. THEOREM.

Two triangles are equal, which have the three sides of the one, respectively equal to the three sides of the other.

Let the side ED=BA, the side EF-BC, and the side DF-AC; then will the angle D=A, the angle E-B, and the angle F-C.

For, if the angle D B

were greater than A, while the sides ED, DF, were equal to BA, AC, each to each, it would follow by the last proposition, that the side EF must be greater than BC; and if the angle D were less than A, it would follow, that the side EF must be less than BC: but EF is equal to BC, by hypothesis; therefore, the angle D can neither be greater nor less than A; hence it must be equal to it. In the same manner it may be shown that the angle E is equal to B, and the angle F to C; hence the two triangles are equal (Def. 34.) Scholium. It may be observed that the equal angles lie opposite the equal sides: thus, the equal angles D and A, lie opposite the equal sides EF and BC.

PROPOSITION XIV. THEOREM.

The angles opposite the equal sides of an isosceles triangle are equal.

Let the sides AB, AC, of the triangle ABC be equal, then will the angle C be equal to the angle B.

For, let AD be the line bisecting the angle A; then, in the two triangles ABD, ACD, two sides. AB, AD, and the included angle in the one, are equal to the two sides AC, AD, and the included angle in the other; hence the angle B is equal to the angle C (Prop. VIII. Cor. 2.)

Cor. 1. It also follows (Prop. VIII. Cor. 2.) that BD

is equal to CD, and that the angle ADB is equal to the angle ADC; therefore, the line bisecting the vertical angle of an isosceles triangle bisects the base at right angles; and conversely, the line bisecting the base of an isosceles triangle at right angles, bisects also the vertical angle.

Cor. 2. Every equilateral triangle is also equiangular.

PROPOSITION XV. THEOREM.

Conversely, if two angles of a triangle are equal, the opposite sides are equal, and the triangle is isiosceles.

In the triangle ABC, let the angles ABC, ACB, be equal; then, if it be supposed that one of the opposite sides, as AB, is longer than the other AC, let BD be equal to AC; then the triangle DCB is obviously less than the triangle ABC. But, since CB, BD, and the included angle, are equal to BC, CA, and the included angle, by hypothesis, it follows that the same triangles are equal (Prop. VIII.), which is impossible; therefore AB cannot be longer than AC, and in a similar manner it may be shown that AC cannot be longer than AB; therefore these two sides are equal.

In

B

Cor. Therefore every equiangular triangle is equilateral.

PROPOSITION XVI. THEOREM.

any triangle the greater angle is opposite the longer side; and conversely, the longer side is opposite the greater angle.

First, Let the angle C be greater than the angle B; then will the side AB, opposite C, be greater than AC, opposite B.

For, make the angle BCP-B.

A

D

Then, in the

triangle CDB, we shall have CD=BD (Prop. XV.) Now, the side AC<AD+CD; but AD+CD=

AD+DB AB; therefore AC<AB.

Secondly, Suppose the side AB>AC; then will the angle C, opposite to AB, be greater than the angle B, opposite to AC.

For, if the angle CB, it follows, from what has just been proved, that AB AC; which is contrary to the hypothesis. If the angle CB, then the side AB-AC (Prop. XV.), which is also contrary to the supposition. Therefore, when AB> AC, the angle C must be greater than B.

PROPOSITION VII. THEOREM.

The perpendicular drawn from a point to a right line is shorter than any other line drawn thereto from the same point; and those lines which meet the proposed line at equal distances from the perpendicular are themselves equal, and the more remote from the perpendicular the point of meeting is, the longer is the line drawn.

Let A be the given point, DE the given line, AB the perpendicular, and AD, AC, AE, the oblique lines.

Produce the perpendicular AB till BF is equal to AB, and draw FC, FD.

First. The triangle BCF is equal to the triangle BCA, for they have the right angle CBF CBA, the side CB common, and

the side BF BA; hence the third sides, CF and CA, are equal (Prop. VIII. Cor. 2.) But ABF, being a right line, is shorter (Def. 8.) than ACF, which is a broken line; therefore, AB, the half of ABF, is shorter than AC, the half of ACF; hence, the perpendicular is shorter than any oblique line.

Secondly. Let us suppose BC-BE; then will the triangle CAB be equal to the triangle BAE; for BC-BE, the side AB is common, and the angle CBA ABE; hence the sides AC and AE are equal (Prop. VIII. Cor. 2.): therefore, two oblique lines, equally distant from the perpendicular, are equal.

Thirdly. In the triangle DFA, the sum of the lines AC, CF, is less than the sum of the sides AD, DF (Prop. XI.); therefore AC, the half of the line ACF, is shorter than AD, the half of the line ADF: hence, the oblique line, which is farther from the perpendicular, is longer than the one which is nearer.

Cor. 1. The perpendicular measures the shortest distance of a point from a line. Hence only one perpendicular can be drawn from a point to a given line.

Cor. 2. From the same point to the same straight line, only two equal straight lines can be drawn; for, if there could be more, we should have at least two equal oblique lines on the same side of the perpendicular, which is impos

sible.

If a perpendicular be drawn from the middle point of a right

line, every point in the perpendicular will be equidistant from the extremities of the line; and every point without the perpendicular, or the perpendicular produced, will be unequally distant from those extremities.

Let AB be the given right line, C the middle point, and ECF the perpendicular.

First. Since AC-CB, the two oblique lines AD, DB, are equally distant from the perpendicular, and therefore equal (Prop. XVII.) So, likewise, are the two oblique A lines AE, EB, the two AF, FB, and so on. Therefore every point in the perpendicular is equally distant from the extremities A and B.

B

Secondly. Let I be a point out of the perpendicular. If IA and IB be drawn, one of these lines will cut the perpendicular in D; from which, drawing DB, we shall have BD

DA. But the straight line IB is less than ID+DB, and ID+DB=ID+DA-IA; therefore, IBIA; hence, every point out of the perpendicular, is unequally distant from the extremities A and B.

Cor. If a straight line have two points D and F, equally distant from the extremities A and B, it will be perpendicular to AB at the middle point C.

PROPOSITION XIX. THEOREM.

Two right angled triangles are equal, when the hypothenuse and one side in one triangle are respectively equal to the hypothenuse and a side in the other.

BC=EF, the angle BAC=D, and the angle C=F.

If the side BC is equal to EF, the like angles of the two triangles are equal (Prop. XIII. Sch.) Now, if it be possible, suppose these two sides to be unequal, and that BC is the greater.

On BC take BG=EF, and draw AG. Then, in the two triangles BAG, DEF, the angles B and E are equal, being right angles, the side BA=ED by hypothesis, and the side BG=EF by construction: consequently, AG=DF (Prop.