If a straight line meets another straight line, the sum of the adjacent angles is equal to two right angles.

Let DC meet AB at C: then

is the sum of the angles DCA and DCB equal to two right angles.

At C, let CE be drawn perpendicular to AB (Post. 6); then, by

definition (D. 12), the angles ECA

E

and ECB are both right angles, and consequently, their sum is equal to two right angles.

The angle DCA is equal to the sum of the angles ECA and ECD (A. 9); hence,

But,

DCA + DCB ECA + ECD + DCB;

ECD + DCB is equal to ECB (A. 9); hence,

DCA + DCB = ECA + ECB.

The sum of the angles ECA and ECB, is equal to two right angles; consequently, its equal, that is, the sum of the angles DCA and DCB, must also be equal to two right angles; which was to be proved.

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

Cor. 2. The sum of the angles BAC, CAD, DAE, EAF, formed about a given point on the same side of a straight line BF, is equal to two right angles. right angles. For, their sum is equal to 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.

1o. ADJACENT ANGLES are 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, ACD, are are adjacent angles.

E

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 proposition 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 are equal.

vertical angles equal.

The sum of the

of the adjacent

angles ACE and ACD, is equal to

two right angles (P. I.): the sum

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,

ACE + ACD = ACE + ECB;

Taking from both the common angle ACE (A. 3), there remains,

ACD ECB.

In like manner, we find,

ACD + ACE ACD + DCB;

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

ACE DCB.

Hence, the proposition is proved.

Cor. 1. If one of the angles about C is a right angle, all of the others are right angles also. For, (P. I., C. 1), each of its adjacent angles is

a right angle; and from the proposition just demonstrated, its opposite angle is also a right angle.

B

Cor. 2. If one line DE, is perpendicular to another AB, then is the second line AB perpendicular to the first DE. For, the angles DCA and DCB are right angles, by 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 perpen dicular 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.

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

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

Let A and B be

be two points

common to two lines: then the lines coincide throughout.

E

B

Between A and B they must 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.

PROPOSITION IV. THEOREM.

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

Let DC meet AC and BC at C, making the sum of the angles DCA and DCB equal to two right angles: then is CB the prolongation of AC.

A

-B
E

For, if not, suppose CE to be the prolongation of AC; then is the sum of the angles DCA and DCE equal to two right angles (P. I.): consequently, we have (A. 1),

DCA DCB DCA + DCE;

Taking from both the common angle DCA, there remains

DCB = DCE,

which is impossible, since a part can not 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 are equal in all respects.

In the triangles ABC and DEF, let AB be equal to DE,