The sign indicates multiplication: thus, Ax B represents the product of A and B. Instead of the sign ×, a point is sometimes employed; thus, A.B is the same thing as Ax B. The same product is also designated without any intermediate sign, by AB; but this expression should not be employed, when there is any danger of confounding it with that of the line AB, which expresses the distance between the points A and B.

The expression Ax (B+C-D) represents the product of A by the quantity B+C-D. If A+B were to be multiplied by A-B+C, the product would be indicated thus, (A+B) × (A-B+C), whatever is enclosed within the curved lines, being considered as a single quantity.

A number placed before a line, or a quantity, serves as a multiplier to that line or quantity; thus, 3AB signifies that the 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 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; √A× B means the square-root of the product of A and B.

Axioms.

1. Things which are equal to the same thing, are equal to each other.

2. If equals be added to equals, the wholes will be equal. 3. If equals be taken from equals, the remainders will be equal.

4. If equals be added to unequals, the wholes will be unequal.

5. If equals be taken from unequals, the remainders will be unequal.

6. Things which are double of the same thing, are equal to each other.

7. Things which are halves of the same thing, are equal to each other.

8. The whole is greater than any of its parts.

9. The whole is equal to the sum of all its parts.

10. All right angles are equal to each other.

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

12. Through the same point, only one straight line can be drawn which shall be parallel to a given line.

13. Magnitudes, which being applied to each other, coincide throughout their whole extent, are equal.

B

PROPOSITION I. THEOREM.

If one straight line meet another straight line, the sum of the two adjacent angles will be equal to two right angles.

A

E

B

Let the straight line DC meet the straight line AB at C, then will the angle ACD + the angle DCB, be equal to two right angles. At the point C, erect CE perpendicular to AB. The angle ACD is the sum of the angles ACE, ECD: therefore ACD+DCB is the sum of the three angles ACE, ECD, DCB: but the first of these three angles is a right angle, and the other two make up the right angle ECB; hence, the sum of the two angles ACD and DCB, is equal to two right angles.

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

Cor. 2. If the line DE is perpendicular to AB, reciprocally, AB will be perpendicular to DE.

For, since DE is perpendicular to AB, the A angle ACD must be equal to its adjacent angle DCB, and both of them must be right angles (Def. 10.). But since ACD is a

C

B

right angle, its adjacent angle ACE must also be a right angle (Cor. 1.). Hence the angle ACD is equal to the angle ACE, Ax. 10.): therefore AB is perpendicular to DE. Cor. 3. The sum of all the successive angles, BAC, CAD, DAE, EAF, formed. C on the same side of the straight line BF, is equal to two right angles; for their sum is equal to that of the two adjacent an-, gles, BAC, CAF.

E

B

A

F

PROPOSITION II. THEOREM.

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

2

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. 11). Sup

pose, however, that on being produced, these lines begin to separate at C, the one becoming CD, the other CE. From the point C draw the line CF, making with AC the right angle ACE. 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 likewise be a right angle. Hence, the angle FCD is equal to the angel FCE (Ax. 10.); which can only be the case when the lines CD and CE coincide: therefore, the straight lines which have two points A and B common, cannot separate at any point, when produced; hence they form one and the same straight line.

PROPOSITION III. THEOREM.

If a straight line meet two other straight lines at a common point, making the sum of the two adjacent angles equal to two right angles, the two straight lines which are met, will form one and the same straight line.

Let the straight line CD meet the two lines AC, CB, at their common point C, making the sum of the two adjacent angles DCA, DCB, equal to A two right angles; then will CB be the prolongation of AC, or AC and CB will form one and the same straight line.

For, if CB is not the prolongation of AC, let CE be that prolongation: then the line ACE being straight, the sum of the angles ACD, DCE, will be equal to two right angles (Prop. I.). But by hypothesis, the sum of the angles 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 angle DCE equal to the angle DCB, which can only be the case when the lines CE and C coincide; hence, AC, CB, form one and the same straight li..

PROPOSITION IV. THEOREM.

When two straight lines intersect each other, the oppor vertical angles, which they form, are equal.

Let AB and DE be two straight A lines, intersecting each other at C; then will the angle ECB be equal to the angle ACD, and the angle ACE to the angle DCB.

B

E

For, since the straight line DE is met by the straight line AC, the sum of the angles ACE, ACD, is equal to two right angles (Prop. I.); and since the straight line AB, is met by the straight line EC, the sum of the angles ACE and ECB, is equal to two right angles: hence the sum ACE+ACD is equal to the sum ACE+ECB (Ax. 1.). Take away from both, the common angle ACE, there remains the angle ACD, equal to its opposite or vertical angle ECB (Ax. 3.).

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 sum of the two angles ACE, ECB, is equal to two right angles; and the sum of the other two, ACD, DCB, is also equal to two right angles: therefore, the sum of the four is equal to four right angles.

B

D

In general, if any number of straight lines CA, CB, CD, &c. meet in a point C, the sum of all the successive 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 two lines pendicular to each other, the same space would be occupied by the four right angles, as by the successive angles ACB, BCD, DCE, ECF, FCA.

per

PROPOSITION V. THEORDM.

T

E

If two triangles have two sides and the included angle of the one, equal to two sides and the included angle of the other, each to each, the two triangles will be equal.

Let the side ED be equal to the side BA, the side DF to the side AC, and the angle D to the angle A; then will the triangle EDF be equal to the triangle BAC. E

D

TB

For, these triangles may be so applied to each other, that they shall exactly coincide. Let the triangle EDF, be placed upon the triangle BAC, so that the point E shall fall upon B, and the side ED on the equal side BA; then, since the angle D is equal to the angle A, the side DF will take the direction AC. But

DF is equal to AC; therefore, the point F will fall on C, and the third side EF, will coincide with the third side BC (Ax. 11.): therefore, the triangle EDF is equal to the triangle BAC (Ax. 13.).

Cor. When two triangles have these three things equal, namely, the side ED=BA, the side DF=AC, and the angle D=A, the remaining three are also respectively equal, namely, the side EF=BC, the angle E=B, and the angle F=C

PROPOSITION VI. THEOREM.

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

Let the angle E be equal to the angle B, the angle F to the angle C, and the inIcluded side EF to the included side BC; then will the triangle EDF be equal to the triangle BAC.

For to apply the one to the other, let the side EF be placed on its equal BC, the point E falling on B, and the point F on C; then, since the angle E is equal to the angle B, the side ED will take the direction BA; and hence 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, falling at the same time in the two straight lines BA and CA, must fall at their intersection A: hence, the two triangles EDF, BAC, coincide with each other, and are therefore equal (Ax. 13.).

Cor. Whenever, in two triangles, these three things are equal, namely, the angle E=B, the angle F-C, and the included side EF equal to the included side BC, it may be inferred that the remaining three are also respectively equal, namely, the angle D=A, the side ED=BA, and the side DF=AC.

Scholium. Two triangles are said to be equal, when being applied to each other, they will exactly coincide (Ax. 13.). Hence, equal triangles have their like parts equal, each to each, since those parts must coincide with each other. The converse of this proposition is also true, namely, that two triangles which have all the parts of the one equal to the parts of the other, each

B*