to each, are equal; for they may be applied to each other, and the equal parts will mutually coincide.

PROPOSITION VII. THEOREM.

The sum of any two sides of a triangle, is greater than the

third side.

Let ABC be a triangle: then will the sum of two of its sides, as AC, CB, be greater than the third side AB.

For the straight line AB is the shortest distance between the points A and B (Def. 3.); hence AC+CB is greater A than AB.

B

C

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

Let any point, as O, be taken within the triangle BAC, and let the lines OB, OC, be drawn to the extremities of either side, as BC; then will OB+OC<BA+AC.

Let BO be produced till it meets the side AC in D: then the line OC is shorter than OD+DCB

(Prop. VII.): add BO to each, and we have BO+OC<BO+ OD+DC (Ax. 4.), or BO+OC<BD+DC.

Again, BD<BA+AD: add DC to each, and we have BD+ DC BA+AC. But it has just been found that BO+OC< BD+DC; therefore, still more is BO+OC<BA+AC.

PROPOSITION IX. THEOREM.

If two triangles have two sides of the one equal to two sides of the other, each to each, and the included angles unequal, the third sides will be unequal; and the greater side will belong to the triangle which has the greater included angle.

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

Make the angle CAGB4 =D; take AG-DE, and draw CG.

The

C

F

triangle GAC is equal to DEF, since, by construction, they have an equal angle in each, contained by equal sides, (Prop. V.); 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, BC>EF.

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 (Ax. 8.).

AA

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. B.

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.

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. Cor.); in the second, CB would be less than EF; but either of these results contradicts the hypothesis: therefore, BAC is greater than EDF.

PROPOSITION X. THEOREM.

If two triangles have the three sides of the one equal to the three sides of the other, each to each, the three angles will also be equal, each to each, and the triangles themselves will be equal.

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.

E

D

A

FB

For, if the angle D 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; therefore 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 (Prop. VI. Sch.).

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 XI. THEOREM.

In an isosceles triangle, the angles opposite the equal sides

are equal.

Let the side BA be equal to the side AC; then will the angle C be equal to the angle B.

A

For, join the vertex A, and D the middle point of the base BC. Then, the triangles BAD, DAC, will have all the sides of the one equal to those of the other, each to each; for BA is equal to AC, by hypothesis; AD is common, and BD is equal to DC by construction: therefore, by the last proposition, the angle B is equal to the angle C.

D

C

Cor. An equilateral triangle is likewise equiangular, that is to say, has all its angles equal.

Scholium. The equality of the triangles BAD, DAC, proves also that the angle BAD, is equal to DAC, and BDA to ADC, hence the latter two are right angles; therefore, the line drawn from the vertex of an isosceles triangle to the middle point of its base, is perpendicular to the base, and divides the angle at the vertex into two equal parts.

In a triangle which is not isosceles, any side may be assumed indifferently as the base; and the vertex is, in that case, the vertex of the opposite angle. In an isosceles triangle, however

that side is generally assumed as the base, which is not equal to either of the other two.

PROPOSITION XII. THEOREM.

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

Let the angle ABC be equal to the angle ACB; then will the side AC be equal to the side AB.

For, if these sides are not equal, suppose AB to be the greater. Then, take BD equal to AC, and draw CD. Now, in the two triangles BDC, BAC, we have BD-AC, by construction; the angle B equal to the angle ACB, by hypothesis; B4 and the side BC common: therefore, the two

D

A

angles, BDC, BAC, have two sides and the included angle in the one, equal to two sides and the included angle in the other, each to each: hence they are equal (Prop. V.). But the part cannot be equal to the whole (Ax. 8.); hence, there is no inequality between the sides BA, AC; therefore, the triangle BAC is isosceles.

PROPOSITION XIII. THEOREM.

The greater side of every triangle is opposite to the greater angle; and conversely, the greater angle is opposite to the greater side.

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

For, make the angle BCD=B. Then, in the triangle CDB, we shall have CD=BD (Prop. XII.). Now, the side AC<AD+CD; but AD+CD= AD+DB=AB: therefore AC<AB.

D

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. It the angle C-B, then the side AB=AC (Prop. XÍÏ.); which is also contrary to the supposition. Therefore, when AB>AC, the angle C must be greater than B.

PROPOSITION XIV. THEOREM.

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

Let A be the point, and DE the given line.

A

BE

F

Let us suppose that we can draw two perpendiculars, AB, AC. Produce either of them, as AB, till BF is equal to AB, and D draw FC. Then, the two triangles CAB, CBF, will be equal: for, the angles CBA, and CBF are right angles, the side CB is common, and the side AB equal to BF, by construction; therefore, the triangles are equal, and the angle ACB=BCF (Prop. V. Cor.). But the angle ACB is a right angle, by hypothesis; therefore, BCF must likewise be a right angle. But if the adjacent angles BCA, BCF, are together equal to two right angles, ACF must be a straight line (Prop. III.): from whence it follows, that between the same two points, A and F, two straight lines can be drawn, which is impossiblo (Ax. 11). hence, two perpendiculars cannot be drawn from the same point to the same straight line.

Scholium. At a given point C, in the line AB, it is equally impossible to erect two perpendiculars to that line. For, if CD, CE, were those two perpendiculars, the angles BCD, BCE, would both be right angles hence they would be equal (Ax. 10.); and

E

C

B

the line CD would coincide with CE; otherwise, a part would be equal to the whole, which is impossible (Ax. 8.).

PROPOSITION XV. THEOREM.

If from a point without a straight line, a perpendicular be let fall on the line, and oblique lines be drawn to different points : 1st, The perpendicular will be shorter than any oblique line. 2d, Any two oblique lines, drawn on different sides of the perpendicular, cutting off equal distances on the other line, will be equal.

3d, Of two oblique lines, drawn at pleasure, that which is farther from the perpendicular will be the longer.