the sum of the angles EAB and EAF; which, from the proposition just demonstrated, is equal to two right angles.

DEFINITIONS.

If two straight lines intersect each other, they form four angles about the point of intersection, which have received different names, with respect to each other.

are

1o. ADJACENT ANGLES those which lie. on the same side of one line, and 'on opposite sides of the other; thus,, ACE and ECB, or ACE and ACD, are adjacent angles.

A

D

B

2o. OPPOSITE, or VERTICAL ANGLES, are those which lie on opposite sides of both lines; thus, ACE and DCB, or ACD and ECB, are opposite angles. From the pro

position just demonstrated, the sum of any two adjacent angles is equal to two right angles.

PROPOSITION II. THEOREM.

If two straight lines intersect each other, the opposite or vertical angles will be equal.

of the adjacent angles ACE and ECB, is also equal to two right angles. But things which are equal to the same thing, are equal to each other (A. 1); hence,

and, taking away the common angle ACD,. we have,

ACE DCB.

Hence, the proposition is proved.

Cor. 1. If one of the angles about all of the others will be right angles also. each of its adjacent angles will

be a right angle; and from the

proposition just demonstrated, its opposite angle will also be a right angle.

A

C

is a right angle,

For, (P. I., C. 1),

D

E

-B

Cor. 2. If one line DE, is perpendicular to another AB, then will the second line AB

be perpendicular to the first DE.

and DCB are right angles, by

For, the angles DCA definition (D. 12); and

from what has just been proved, the angles ACE and BCE are also right angles. Hence, the two lines are mutually perpendicular to each other.

Cor. 3. The sum of all the angles ACB, BCD, DCE, ECF, FCA, that can be formed about a point, is equal to four right angles.

F

B

租

For, if two lines be drawn through the point, mutually perpendicular to each other, the sum of the angles which they form will be equal to four right angles, and it will also be equal to the sum of the given angles (A. 9). Hence, the sum of the given angles is equal to four right angles.

PROPOSITION III. THEOREM.

If two straight lines have two points in common, they will coincide throughout their whole extent, and form one and the same line.

coincide (A. 11). Suppose, now, that they begin to separate at some point C, beyond AB, the one becoming ACE, and the other ACD. If the lines do separate at C, one or the other must change direction at this point; but this is contradictory to the definition of a straight line (D. 4): hence, the supposition that they separate at any point is absurd. They must, therefore, coincide throughout; which was to be proved.

Cor. Two straight lines can intersect in only one point.

NOTE. The method of demonstration employed above, is called the reductio ad absurdum. It consists in assuming an hypothesis which is the contradictory of the proposition to be proved, and then continuing the reasoning until the assumed hypothesis is shown to be false. Its contradictory is thus proved to be true. This method of demonstration is often used in Geometry.

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

Let DC meet AC and BC

of the at C, making the sum angles DCA and DCB equal to two right angles: then will CB be the prolongation of AC.

D

A

-B

E

For, if not, suppose CE to be the prolongation of AU,

then will the sum of the angles

DCA and DCE be

equal to two right angles (P. I.): have (A. 1),

We shall, consequently,

DCA + DCB = DCA + DCE;

Taking from both the common angle DCA, there re

which is impossible, since a part cannot be equal to the whole (A. 8). Hence, CB must be the prolongation of AC; which was to be proved.

PROPOSITION V. THEOREM.

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 triangles will be equal in all their parts.

In the triangles ABC and DEF, let AB be equal

to DE, AC to DF, and the angle A to the angle D: then will the triangles be equal in all their parts.

For, let ABC be applied to DEF, in such a manner that the angle A shall coincide with the angle D, the side AB taking

the direction DE, and

AA

B

E

Then, because AB is

the side AC the direction DF. equal to DE, the vertex B will coincide with the vertex E; and because AC is equal to DF, the vertex C will coincide with the vertex F; consequently, the side BC will coincide with the side EF (A. 11). The two triangles, therefore, coincide throughout, and are consequently equal in all their parts (I., D. 14); which was to be proved.

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 triangles will be equal in all their parts..

will the triangles be equal in all their parts.

For, let ABC be applied to DEF in such a manner that the angle B shall coincide with the angle E, the side