PROPOSITION !. THEOREM.

All ight angles are equal to each other.

D K

BE

HI

G

Le. the straight line CD be perpendicular to AB, and GII to EF; then, by definition 10, each of the angles ACD, BCD, EGH, FGH, will be a right angle; and it is to be proved that the angle ACD is equal to the angle EGH. Take the four straight lines AC, CB, EG, GF, all equal tc each other; then will the line AB be equal to the line EF (Axiom 2). Let the line EF be applied to the line AB so that the point E may be on A, and the point F on B; then will the lines EF, AB coincide throughout; for otherwise two different straight lines might be drawn from one point to another, which is impossible (Axiom 11). Moreover, since the line EG is equal to the line AC, the point G will fall on the point C; and the line EG, coinciding with AC, the line GH will coincide with CD. For, if it could have any other position, as CK, then, because the angle EGH is equal to FGH (Def. 10), the angle ACK must be equal to BCK, and therefore the angle ACD is less than BCK. Bu BCK is less than BCD (Axiom 9); much more, then, is ACD .ess than BCD, which is impossible, because the angle ACD is equal to the angle BCD (Def. 10); therefore, GH can not but coincide with CD, and the angle EGH coincides with the angle ACD, and is equal to it (Axiom 8). Therefore, all right angles are equal to each other.

PROPOSITION II. THEOREM.

The angles which one straight line makes with another, up › one side of it, are either two right angles, or are together equs to two right angles.

Let the straight line AB make with CD, upon one side of it, the angles ABC, ABD; these are either two right angles, or are together equal to two right angles.

For if the angle ABC is equal to ABD, ch of them is a right angle (Def 10); but

C

B

B

D

if not, suppose the line BE to be drawn from the point B, perpendicular to CD; then will each of the angles CBE, DBE be a right angle. Now the angle CBA is equal to the sum of the two angles CBE, EBA. To each of these equals add the angle ABD; C then the sum of the two angles CBA, ABD will be equal tc the sum of the three angles CBE, EBA, ABD (Axiom 2). Again, the angle DBE is equal to the sum of the two angles DBA, ABE. Add to each of these equals the angle EBC; then will the sum of the two angles DBE, EBC be equal to the sum of the three angles DBĂ, ABE, EBC. Now things that are equal to the same thing are equal to each other (Axiom 1); therefore, the sum of the angles CBA, ABD is equal to the sum of the angles CBE, EBD. But CBE, EBD are two right angles; therefore ABC, ABD are together equal to two right angles. Therefore, the angles which one straight line, &c.

Corollary 1. If one of the angles ABC, ABD is a right angle, the other is also a right angle.

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

For, because DE is perpendicular to AB, A the angle DCA must be equal to its adjacent angle DCB (Def. 10), and each of them must

D

E

B

be a right angle. But since ACD is a right angle, its adjacent angle, ACE, must also be a right angle (Cor. 1). Hence the angle ACE is equal to the angle ACD (Prop. I.), and AB is perpendicular to DE.

Cor. 3. The sum of all the angles BAC, CAD, DAE, EAF, formed on the same side of the line BF, is equal to two right c. angles; for their sum is equal to that of the two adjacent angles BAD, DAF

B

PROPOSITION III. THEOREM (Converse of Prop. II.).

E

F

If, at a point in a straight line, two other straight lines, upon ine opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines are in one and the same straight line.

At the point B, in the straight line AB, let the two straight lines BC, BD, upon the opposite sides of AB, make the adja cent angles, ABC, ABD, together equal to two righ angles

then will BD be in the same straight ine with CB.

B

E

For, if BD is not in the same straight line with CB, let BE be in the same straight line with it; then, because the straight line CBE is met by the straight C line AB, the angles ABC, ABE are together equal to two right angles (Prop. II.). But, by hypothesis, the angles ABC ABD are together equal to two right angles; therefore, the sum of the angles ABC, ABE is equal to the sum of the an gles ABC, ABD. Take away the common angle ABC, and the remaining angle ABE, is equal (Axiom 3) to the remaining angle ABD, the less to the greater, which is impossible. Hence BE is not in the same straight line with BC; and in like manner, it may be proved that no other can be in the same straight line with it but BD. Therefore, if at a point, &c

PROPOSITION IV. THEOREM.

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

Let there be two straight lines, having the points A and B in common; these lines will coincide throughout their whole

extent.

F:

A B C

D

It is plain that the two lines must coincide between A and B, for otherwise there would be two straight lines between A and B, which is impossible (Axiom 11). Suppose, however, that, on being produced, these lines begin to diverge at the point C, one taking the direction CD, anc. the other CE. From the point C draw the line CF at rignt angles with AC; then, since ACD is a straight line, the angle FCD is a right angle (Prop. II, Cor. 1); and since ACE is a straight line, the angle FCE is also a right angle; therefore (Prop. I.), the angle FCE is equal to the angle FCD, the less to the greater, which L absurd. Therefore, two straight lines which have, &c.

PROPOSITION V. 'THEOREM.

If two straight lines cut one another, the vertical or opposi angles are equal.

I et the two straigh, lines, AB, CD, cut one another in the

point E; then will the angle AEC be equal to the angle BED, and the angle AED to the angle CEB.

E

For the angles AEC, AED, which the A straight line AE makes with the straight line CD, are together equal to two right angles (Prop. II.); and the angles AED, DEB, which the straight line DE makes with the straight line AB, are also together equal to two right angles; therefore, the sum of the two angles AEC, AED is equal to the sum of the two angles AED, DEB. Take away the common angle AED, and the remaining angle, AEC, is equal to the remaining angle DEB (Axiom 3). In the same manner, it may be proved that the angle AED is equal to the angle CEB. Therefore, if two straight lines, &c.

Cor. 1. Hence, if two straight lines cut one another, the four angles formed at the point of intersection, are together equal to four right angles.

Cor. 2. Hence, all the angles made by any number of straight lines meeting in one point, are together equal to four right angles.

PROPOSITION VI. THEOREM.

If two triangles have two sides, and the included angle of tre une, equal to two sides and the included angle of the other, each to each, the two triangles will be equal, their third sides will be equal, and their other angles will be equal, each to each.

Let ABC, DEF be two triangles, having the side AB equal to De, and AC to DF, and also the angle A equal to the angle D; then will the triangle ABC be equal to the triangle DEF

B

A

CE

F

For, if the triangle ABC is applied to the triangle DEF, so that the point A may be on D, and the straight line AB upon DE, the point B will coincide with the point E, because AB is equal to DE; and AB, coinciding with DE, AC will coincide with DF, because the angle A is equal to the angle D. Hence, also, the point C will coincide with the point F, because AC is equal to DF. But the point B coincides with the point E; therefore the base BC will coincide with the base EF (Axiom 11), and will be equal to it. Hence, also, the whole triangle ABC will coin cide with the whole triangle DEF, and will be equal to it

B

:

and the reinaining angles of the one, will coincide with the remaining angles of the other, and be equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to the an ge DFE. Therefore, if two triangles, &c.

PROPOSITION VII. 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, the other sides will be equal, each to each, and the third angle of the one to the third angle of the other.

Let ABC, DEF be two triangles having the angle B equal to E, the angle C equal to F, and the inclu

ded sides BC, EF equal to each other; then will the B

.A

triangle ABC be equal to the triangle DEF.

CE

F

For, if the triangle ABC is applied to the triangle DEF, su that the point B may be on E, and the straight line BC upon EF, the point C will coincide with the point F, because BC is equal to EF. Also, since the angle B is equal to the angle E, the side BA will take the direction ED, and therefore the point A will be found somewhere in the line DE. And because the angle C is equal to the angle F, the line CA will take the direction FD, and the point A will be found somewhere in the line DF; therefore, the point A, being found at the same time in the two straight lines DE, DF, must fall at their intersection, D. Hence the two triangles ABC, DEF coincide throughout, and are equal to each other; also, the two sides AB, AC are equal to the two sides DE, DF, each to each, and the angle A to the angle D. Therefore, if two riangles, &c.

PROPOSITION VIII. THEOREM.

Any side of a triangle is less than the sum of the other two

Let ABC be a triangle; any one of its sides is less than the sum of the other two, viz.: the side AB is less than the sum of AC and BC; BC is less than the sum of AB and AC; and AC is less than the sum of AB B and BC.

A