line AB is taken three times; A signifies the half of the angle A.

The square of the line AB is designated by AB2; its cube by AB3. What is meant by the square and the cube of a line will be explained in its proper place.

The sign indicates a root to be extracted; thus, √2 means the square-root of 2; AxB means the square-root of the product of A and B, or the mean proportional between them.

Axioms.

1. Two quantities equal to a third are equal to each other. 2. The whole is greater than its part.

3. The whole is equal to the sum of the parts into which it has been divided.

4. From one point to another, only one straight line can be drawn.

5. Two magnitudes, lines, surfaces, or solids, are equal, if, when applied to each other, they coincide throughout their whole

extent.

PROPOSITION I. THEOREM.

All right angles are equal to each other.

LET the straight line CD be perpendicular to AB, and GH to EF; the angles ACD and EGH will be equal to each other.

D

H

G

F

B

Take the four distances CA, CB, GE, GF all equal; the distance AB will be equal to the distance EF, and the line EF being placed on AB, so that the point E falls on A, the point F may be made to fall on B. Those two lines will thus coincide entirely; for otherwise there would be two E straight lines extending from A to B, which (Ax. 4.) is impossible and hence G, the middle point of EF, will fall on C, the middle point of AB. The side GE being thus applied to CA, the side GH must fall on CD. For suppose, if possible, that it falls on a line CK different from CD: then, since by hypothesis (def. 10.) the angle EGH-HGF, ACK would in that case be equal to KCB. But the angle ACK is greater than

:

ACD; and KCB is smaller than BCD, and by hypothesis ACD=BCD; hence ACK is greater than KCB. Therefore the line GH cannot fall on a line CK different from CD; therefore it falls on CD, and the angle EGH on ACD; therefore all right angles are equal to each other.

PROPOSITION II. THEOREM.

Every straight line CD, which meets another AB, makes with it two adjacent angles ACD, BCD, the sum of which is equal to two right angles.

E

AT the point C, erect CE perpendicular to AB. The angle ACD is the sum of the angles ACE, ECD; therefore ACD+BCD will be the sum of the three ACE, ECD, BCD: but the first of those three is a right angle, and the other two together make up the right angle BCE; hence the sum of the two angles ACD and BCD is equal to two right angles.

A

B

Cor. 1. If one of the angles ACD, BCD is right, the other must be right also.

Cor. 2. If the line DE is perpendicular to AB, in like manner AB will be perpendicular to DE.

For, since DE is perpendicular to AB, the angle ACD must be equal to its adjacent one DCB, and both of them must be right. But since ACD is a right angle, its adjacent

K

B

one ACE must also be right: hence the angle ACE=ACD; hence AB is perpendicular to DE.

Cor. 3. All the consecutive angles BAC, CAD, DAE, EAF formed on the same side of a straight line BF, are equal, when taken together, to two right angles; because their sum is equal to that of the two adjacent angles, BAC, CAF.

B

PROPOSITION III. THEOREM.

D

E

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

LET A and B be the two common points. In the first place, it is evident that the two lines must coincide entirely between A and B, for otherwise there would be two straight lines between A and B, which is impossible (Ax. 4.). Suppose, however, that on being

A

B

F

produced, those lines begin to separate at C, the one becoming CD, the other CE. From the point C draw the line CF, making with CA the right angle ACF. Now since ACD is a straight line, the angle FCD will be right (Prop. 2. cor. 1.); and since ACE is a straight line, the angle FCE will likewise be right. But the part FCE cannot be equal to the whole FCD; hence the straight lines which have two points A and B common, cannot separate in any point of their production; hence they form one and the same straight line.

PROPOSITION IV. THEOREM.

If two adjacent angles, ACD, DCB, are together equal to two right angles, the two exterior sides, AC, CB, will lie in the same straight line.

FOR if CB is not the production of AC, let CE be that production: then the line ACE being straight, the sum of the angles ACD, DCE, will (Prop. A 2.) be equal to two right angles. But, by hypothesis, the sum of the angles

B

E

ACD, DCB is also equal to two right angles: therefore, ACD+ DCE must be equal to ACD+DCB; and taking away the angle ACD from each, there remains the part DCB equal to the whole DCE, which is impossible; therefore, CB is the production of AC.

PROPOSITION V. THEOREM.

Whenever two straight lines AB, DE, intersect each other, the opposite or vertical angles, which they form, are equal.

FOR, since DE is a straight line, the A sum of the angles ACD, ACE is equal to two right angles; and since AB is a straight line, the sum of the angles ACE, BCE is also equal to two right angles:

hence the sum ACD+ACE is equal to the sum ACE+BCE. Take away from both the same angle ACE; there remains the angle ACD, equal to its opposite or vertical angle BCE.

It may be shewn, in the same manner, that the angle ACE is equal to its opposite angle BCD.

Scholium. The four angles formed about a point by two straight lines which intersect each other, are together equal to four right angles: for the two angles ACE, BCE, taken to

gether, are equal to two right angles; and the other two, ACD, BCD have the same value.

In general, if any number of straight lines CA, CB, CD, &c. meet in a point C, the sum of all the consecutive angles ACB, BCD, DCE, ECF, FCA, will be equal to four right angles: for if four right angles were formed about the point C, by means of two lines perpendicular to each other, the same space would be occupied, either

B

by the four right angles, or by the successive angles ACB, BCD, DCE, ECF, FCA.

PROPOSITION VI. THEOREM.

D

Two triangles are equal when an angle, and the two sides which contain it in the one, are respectively equal to an angle and the two sides which contain it in the other.

For those triangles may be applied to each other, so that they shall perfectly coincide. If the side DE be placed on its equal AB, the point D will fall on A, and the point E on B: and, since the angle D is equal to the angle A, when the side DE is placed on AB, the side DF will take the direction AC. Besides, DF is equal to AC; therefore, the point F will fall on C, and the third side EF will exactly cover the third side BC; therefore (Ax. 5.), the triangle DEF is equal to the triangle ABC.

Cor. When, in two triangles, these three things are equal, namely, the angle A=D, the side AB-DE, and the side AC=DF, the other three are equal also, namely, the angle B-E, the angle C-F, and the side BC=EF.

PROPOSITION VII. THEOREM.

Two triangles are equal, if two angles and the interjacent side of the one are equal to two angles and the interjacent side of the other.

LET the side BC (see the last figure) be equal to the side EF, the angle B to the angle E, and the angle C to the angle F; then will the triangle DEF be equal to the triangle ABC.

For, to bring about the superposition, let EF be placed on its equal BC; the point E will fall on B, and the point F on

C.

And, since the angle E is equal to the angle B, the side ED will take the direction BA; therefore, the point D will be found somewhere in the line BA. In like manner, since the angle F is equal to the angle C, the line FD will take the direction CA, and the point D will be found somewhere in the line CA. Hence, the point D, occurring at the same time in the two straight lines BA and CA, must fall on their intersection A; hence the two triangles ABC, DEF, coincide with each other, and are perfectly equal.

Cor. Whenever, in two triangles, these three things are equal, namely, BC=EF, B=E, C=F, it may be inferred that the other three are equal also, namely, AB=DE, AC=. DF, A=D.

PROPOSITION VIII. THEOREM.

In every triangle, any side is less than the sum of the other two.

For the line BC, for example (see the preceding figure), is the shortest distance from B to C; therefore, BC is less than BA+AC.

PROPOSITION IX. THEOREM.

If, from a point O assumed within the triangle ABC, straight lines OB, OC, be drawn to the extremities of a side BC, the sum of these straight lines will be less than that of the two other sides AB, AC.

LET BO be produced till it meet the side AC in D. The line OC (Prop. 8.) is shorter than OD+DC: add BO to each, and we have BO+ OC BO+OD+DC, or BO+OCZBD+DC.

In like manner, BD BA+AD: add DC to each; and we have BD+DC BA+AC. But we have just found BO+OC<BD+DC; therefore, still more is BO+OC_BA+AC.