BOOK I.

THEOREM 1.

When one line meets another, the sum of the two angles which it makes on the same side of the other line, is equal to two right angles. Let AB meet CD; then we are to demonstrate that the two angles ABD+ABC= two right angles.

C

E

A

B

D

If AB does not incline on either side of CD and the angle ABD ABC, then these angles are right angles by definition 9. But if these angles are unequal, conceive the dotted line, BE, drawn from the point B, so as not to incline on either side; then by the definition, the angles CBE and EBD are right angles; but the angles CBA+ABD make the same sum, or fill the same angular space, as the two angles CBE and EBD; therefore, CBA+ABD=two right angles. Q. E. D. *

Cor. 1. Hence, all the angles which can be made at any point B, by any number of lines on the same side of the right line CD, are, when taken all together, equal to two right angles.

Cor. 2. And, as all the angles that can be made on the other side of the line CD are also equal to two right angles, therefore all the angles that can be made quite round a point B, by any number of lines, are equal to four right angles.

Cor. 3. Hence, also, the whole circumference of a circle, being the sum of the measures of all the angles that can be made about the center F, (def. 8), is the measure of four right angles; consequently, a semicircle, or 180 degrees, is

F

the measure of two right angles; and a quadrant, or 90 degrees, the measure of one right angle.

The initials of a Latin phrase, meaning "which was to be demonstrated."

If one straight line meets two other straight lines at a common point, forming two angles, which together make two right angles, the two straight lines are one and the same line.

If AB meets the two lines DB and BC at the common point B, and the two angles DBA+ABC two right angles, then we are to demonstrate that DB and BC form one and the same straight line.

If DB and BC are not in the

But by (hy.)

By subtraction

ABD+ABE=2R (2 R indicates two

ABD+ABC=2R right angles.)

ABE-ABC=0

That is, the angle CBE is zero; and DBC is a continued line; or BC falls on BE.

THEOREM 3.

Q. E. D.

If two straight lines intersect each other, the opposite vertical angles are equal.

If AB and CD intersect each other at E, we are to demonstrate that the angle AEC equals its opposite angle DEB, and AED=CEB.

A

E

B

As AEB is a right line, EA is exactly in the opposite direction from EB; and for the same reason EC is opposite in direction from ED; therefore, the difference in direction between EA and EC is equal to the difference in direction between EB and ED; or by (def. 7), the angle AEC=DEB. In the same manner we can show that the angle AED=CEB. Q. E. D.

Otherwise Let AEC=2, AED=y, and DEB=x; then we are to show that x=z. As AB is a right line, and DE falls upon it, x+y=2R .z+y=2R

we have, by (th. 1),

Also,

By subtraction,

By transposition,

.x-2=0

x=2 Q. E. D.

THEOREM 4.

If a straight line falls across two parallel straight lines, the sum of the two interior angles on the same side of the crossing line is equal to two right angles.

Let AB and CD be two parallel lines, and EF running across them; then we are to demonstrate that the angle BGH+GHD=2R.

Because GB and HD are parallel, C

they are equally inclined to the line

EF, or have the same difference of

direction from that line: Therefore FGB=GHD. To each of these equals add the JBGH.

Then

FGB+BGH=GHD+BGH.

But by (th. 1) the first member of this equation is equal to two right angles that is, the two interior angles GHD and BGH are together equal to two right angles. Q. E. D.

THEOREM 5.

If a straight line falls across two parallel straight lines, the interior alternate angles are equal; and also the opposite exterior angles. On the supposition that AB and CD are parallel, (see last figure), and EF falls across them, we are to demonstrate

1st. That the JAGH the alternate GHD.

2d. That AGF=EHD; or FGB=CHE.

By the definition of parallel lines we have

2d. The

FGB GHD

But FGB=AGH (th. 3)

Hence AGH GHD (ax. 1) Q. E. D.

FGB=GHD. But GHD-CHE (th. 3); therefore, FGB CHE. In the same manner we prove that AGF is equal to EHD. Q. E. D.

If a straight line falls across two parallel straight lines, the exterior angles are equal to the interior opposite angles on the same side of the crossing line.

If AB and CD are parallel, (see last figure), and EF crosses them, then we are to prove that the exterior

FGB=GHD

AGH-FGB (th. 3)

AGF CHG

AGH GHD (th. 5)

Hence FGB=GHD (ax. 1)

Q. E. D.

In the same manner we prove that AGF=CHG.

THEOREM 7.

If a straight line falls across two other straight lines, and makes the sum of the two interior angles on the same side equal to two right angles, the two straight lines must be parallel.

Let EF be the line falling across the lines AB and CD, making the two angles BGH+GHD to two right angles; then we are to demonstrate that AB and CD must be parallel.

As EF is a right line, and BA meets it, the two angles (th. 1)

By (hy.).

By subtraction,

FGB+BGH=2R

GHD+BGH=2R

FGB-GHD=0. That is, there is no differ

ence in the direction of GB and HD from the same line EF; but when there is no difference in the direction of lines (def. 13) the lines are parallel; therefore, AB and CD are parallel. Q. E. D.

THEOREM 8.

Parallel lines can never meet, however far they may be produced. If the lines AB and CD (see last figure) should meet at any distance on either side of EF, they would there form an angle; and if they formed an angle they would not run in the same direction; and not running in the same direction, they would not be parallel; but by (hy.) they are parallel; therefore they cannot meet. Q. E. D.

THEOREM 9.

If two straight lines are parallel to a third, they are parallel to each other.

If AB is parallel to EF, and CD also parallel to EF, then we are to show that AB is parallel to CD.

Because AB and EF are parallel, E

they make equal angles with the

line HG (def. 13, 2); and because

D

CD and EF are parallel, those two lines make equal angles with the line HG.

Hence AB and CD, making equal angles with another line that falls across them, they are therefore parallel (def. 7). Q. E. D.

THEOREM 10.

If two angles have their sides parallel, the two angles will be equal.

Scholium. When AH extends in the opposite direction, it is still parallel to BF; but the angle then is the supplemental angle to DBF; that is, equal to FBG.