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

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

D

B

D

B

For if the angle CBA is equal to the angle ABD, each of them is a right angle. [Definition 10. But if not, from the point B draw BE at right angles to CD; [I. 11. therefore the angles CBE, EBD are two right angles. [Def.10. Now the angle CBE is equal to the two angles CBA, ABE; to each of these equals add the angle EBD;

therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. [Axiom 2. Again, the angle DBA is equal to the two angles DBE, EBA;

to each of these equals add the angle ABC;

therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC.

[Axiom 2. But the angles CBE, EBD have been shown to be equal to the same three angles.

Therefore the angles CBE, EBD are equal to the angles DBA, ABC.

But CBE, EBD are two right angles;

[Axiom 1.

therefore DBA, ABC are together equal to two right angles.

Wherefore, the angles &c. Q.E.D.

PROPOSITION 14. THEOREM.

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

At the point B in the straight line AB, let the two straight lines BC, BD, on the opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles: BĎ shall be in the same straight line with CB. For if BD be not in

the same straight line with CB, let BE be in the same straight line with it.

Then because the straight line AB makes with the straight line CBE, on one side of it, the angles ABC, ABE, these angles are to

B

gether equal to two right angles.

[I. 13.

But the angles ABC, ABD are also together equal to two

right angles. [Hypothesis. Therefore the angles ABC, ABE are equal to the angles ABC, ABD.

From each of these equals take away the common angle ABC, and the remaining angle ABE is equal to the remaining angle ABD, [Axiom 3.

the less to the greater; which is impossible.

Therefore BE is not in the same straight line with CB.

And in the same manner it may be shewn that no other can be in the same straight line with it but BD; therefore BD is in the same straight line with CB. Wherefore, if at a point &c.

Q.E.D.

PROPOSITION 15. THEOREM.

If two straight lines cut one another, the vertical, or opposite, angles shall be equal.

Let the two straight lines AB, CD cut one another at the point E; the angle AEC shall be equal to the angle DEB, and the angle CEB

to the angle AED.

Because the straight line AE makes with the straight line CD the angles CEA, AED, these angles are together equal to two right angles.

A

[I. 13.

Again, because the straight line DE makes with the straight line AB the angles AED, DEB, these also are together equal to two right angles.

[I. 13.

But the angles CEA, AED have been shewn to be together equal to two right angles.

Therefore the angles CEA, AED are equal to the angles AED, DEB.

From each of these equals take away the common angle AED, and the remaining angle CEA is equal to the remaining angle DEB.

[Axiom 3. In the same manner it may be shewn that the angle CEB is equal to the angle AED.

Wherefore, if two straight lines &c. Q.E.D.

Corollary 1. From this it is manifest that, if two straight lines cut one another, the angles which they make at the point where they cut, are together equal to four right angles.

Corollary 2. And consequently, that all the angles made by any number of straight lines meeting at one point, are together equal to four right angles.

PROPOSITION 16. THEOREM.

If one side of a triangle be produced, the exterior angle shall be greater than either of the interior opposite angles.

Let ABC be a triangle, and let one side BC be produced to D: the exterior angle ACD shall be greater than either of the interior opposite angles CBA, BAC.

Bisect AC at E,

[I. 10. join BE and produce it to F, making EF equal to EB, [I. 3. and join FC.

Because AE is equal to EC, and BE to EF; [Constr. the two sides AE, EB are equal to the two sides CE, EF, each to each;

and the angle AEB is equal to the angle CEF,
because they are opposite ver-
tical angles;

[I. 15.
therefore the triangle AEB
is equal to the triangle CEF,
and the remaining angles to
the remaining angles, each to
each, to which the equal sides
are opposite;
[I. 4.
therefore the angle BAE is
equal to the angle ECF.

But the angle ECD is greater than the angle ECF. [Axiom 9.

E

Therefore the angle ACD is greater than the angle BAE. In the same manner if BC be bisected, and the side AC be produced to G, it may be shewn that the angle BCG, that is the angle ACD, is greater than the angle ABC. [I. 15. Wherefore, if one side &c. Q.E.D.

PROPOSITION 17. THEOREM.

Any two angles of a triangle are together less than two right angles.

Let ABC be a triangle: any two of its angles are together less than two right angles.

Produce BC to D.

Then because ACD is the exterior angle of the triangle ABC, it is greater than the interior opposite angle ABC. [I. 16.

To each of these add the angle ACB

Therefore the angles ACD, ACB

are greater than the angles ABC, ACB.

But the angles ACD, ACB are together equal to two right angles.

[I. 13. Therefore the angles ABC, ACB are together less than two right angles.

In the same manner it may be shewn that the angles BAC, ACB, as also the angles CAB, ABC, are together less than two right angles.

Wherefore, any two angles &c. Q.E.D.

PROPOSITION 18. THEOREM.

The greater side of every triangle has the greater angle opposite to it.

Let ABC be a triangle, of which the side AC is greater than the side AB: the angle ABC is also greater than the angle ACB.

Because AC is greater than AB, make AD equal to AB, [I. 3. and join BD.

Then, because ADB is the exterior angle of the triangle BDC, it is greater than the interior opposite angle DCB.

[I. 16.

B

[I. 5.

[Constr.

But the angle ADB is equal to the angle ABD,
because the side AD is equal to the side AB.
Therefore the angle ABD is also greater than the angle
АСВ.

Much more then is the angle ABO greater than the angle
АСВ.

Wherefore, the greater side &c. Q.E.D.

PROPOSITION 19. THEOREM.

[Axiom 9.

The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it.

Let ABC be a triangle, of which the angle ABC is greater than the angle ACB: the side AC is also greater than the side AB.

For if not, AC must be either equal to AB or less than AB. But AC is not equal to AB, for then the angle ABC would be equal to the angle ACB; [1.5. but it is not;

[Hypothesis.

therefore AC is not equal to AB.

Neither is AC less than AB,

for then the angle ABC would be less than the angle

ACB;

but it is not;

[I. 18. [Hypothesis.