PROPOSITION XV. 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 in the point E. Then 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 CD at the point E, the adjacent angles CEA, AED;

these angles are together equal to two right angles. (1. 13.) Again, because the straight line DE makes with AB at the point E, the adjacent angles AED, DEB;

these angles also are equal to two right angles; but the angles CEA, AED have been shewn to be equal to two right angles; wherefore the angles CEA, AED are equal to the angles AED, DEB; take away from each the common angle AED,

and the remaining angle CEA is equal to the remaining angle DEB. (ax. 3.)

In the same manner it may be demonstrated, that the angle CEB is equal to the angle AED.

Therefore, if two straight lines cut one another, &c. Q. E.D.

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

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

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

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

Bisect AC in E, (1. 10.) and join BE;

produce BE to F, making EF equal to BE, (1. 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 CE, EF, each to each, in the triangles ABE, CFE;

and the angle AEB is equal to the angle CEF, because they are opposite vertical angles; (I. 15.) therefore the base AB is equal to the base CF, (1. 4.) and the triangle AEB to the triangle CEF,

and the remaining angles of one triangle to the remaining angles of the other, each to each, to which the equal sides are opposite; wherefore the angle BAE is equal to the angle ECF;

but the angle ECD or ACD is greater than the angle ECF; therefore the angle ACD is greater than the angle BAE or BAC. In the same manner, if the side BC be bisected, and AC be produced to G; it may be demonstrated that the angle BCG, that is, the angle ACD, (I. 15.) is greater than the angle ABC.

Therefore, if one side of a triangle, &c. Q. E.D.

PROPOSITION XVII. THEOREM.

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

Then any two of its angles together shall be less than two right angles.

Produce any side BC to D.

Then because ACD is the exterior angle of the triangle ABC; therefore the angle ACD is greater than the interior and opposite angle ABC; (1. 16.)

to each of these unequals add the angle ACB;

therefore the angles ACD, ACB are greater than the angles ABC, ACB;

but the angles ACD, ACB are equal to two right angles; (I. 13.) therefore the angles ABC, ACB are less than two right angles. In like manner it may be demonstrated,

that the angles BAC, ACB are less than two right angles,
as also the angles CAB, ABC.

Therefore any two angles of a triangle, &c. Q. E.D.

PROPOSITION XVIII. THEOREM.

The greater side of every triangle is opposite to the greater angle.

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

Then the angle ABC shall be greater than the angle ACB.

Since the side AC is greater than the side AB, (hyp.)
make AD equal to AB, (1. 3.) and join BD.

Then, because AD is equal to AB, in the triangle ABD, therefore the angle ABD is equal to the angle ADB, (1. 5.) but because the side CD of the triangle BDC is produced to A, therefore the exterior angle ADB is greater than the interior and opposite angle DCB; (1. 16.)

but the angle ADB has been proved equal to the angle ABD, therefore the angle ABD is greater than the angle DCB; wherefore much more is the angle ABC greater than the angle ACB. Therefore the greater side, &c. Q. E.D.

PROPOSITION XIX. THEOREM.

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 BCA.

Then the side AC shall be greater than the side AB.

A

B

For, if AC be not greater than AB,
AC must either be equal to, or less than AB;
if AC were equal to AB,

then the angle ABC would be equal to the angle ACB; (1. 5.) but it is not equal; (hyp.)

therefore the side AC is not equal to AB.
Again, if AC were less than AB,

then the angle ABC would be less than the angle ACB; (1. 18.) but it is not less, (hyp.)

therefore the side AC is not less than AB; and AC has been shewn to be not equal to AB;

therefore AC is greater than AB. Wherefore the greater angle, &c.

PROPOSITION XX.

Q.E.D.

THEOREM.

Any two sides of a triangle are together greater than the third side.

Let ABC be a triangle.

Then any two sides of it together shall be greater than the third side, viz. the sides BA, AC greater than the side BC;

Produce the side BA to the point D,

make AD equal to AC, (1. 3.) and join DC.
Then because AD is equal to AC, (constr.)

therefore the angle ACD is equal to the angle ADC; (1. 5.) but the angle BCD is greater than the angle ACD; (ax. 9.) therefore also the angle BCD is greater than the angle ADĆ. And because in the triangle DBC,

the angle BCD is greater than the angle BDC,

and that the greater angle is subtended by the greater side; (1. 19.)
therefore the side DB is greater than the side BC;
but DB is equal to BA and AC,

therefore the sides BA and AC are greater than BC.
In the same manner it may be demonstrated,
that the sides AB, BC are greater than CA;
also that BC, CA are greater than AB.
Therefore any two sides, &c. Q.E.D.

PROPOSITION XXI. THEOREM.

If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle; these shall be less than the other two sides of the triangle, but shall contain a greater angle.

Let ABC be a triangle, and from the points B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to a point D within the triangle.

Then BD and DC shall be less than BA and AC the other two sides of the triangle,

but shall contain an angle BDC greater than the angle BAC.

Produce BD to meet the side AC in E.

Because two sides of a triangle are greater than the third side, (1. 20.) therefore the two sides BA, AE of the triangle ABE are greater than BE;

to each of these unequals add EC;

therefore the sides BA, AC are greater than BE, EC. (ax. 4.) Again, because the two sides CE, ED of the triangle CED are greater than DC; (1. 20.)

add DB to each of these unequals;

therefore the sides CE, EB are greater than CD, DB. (ax. 4.) But it has been shewn that BA, AC are greater than BÈ, EC; much more then are BA, AC greater than BD, DC. Again, because the exterior angle of a triangle is greater than the interior and opposite angle; (1. 16.)

therefore the exterior angle BDC of the triangle CDE is greater than the interior and opposite angle CED;

for the same reason, the exterior angle CED of the triangle ABE is greater than the interior and opposite angle BAC; and it has been demonstrated,

1.e. Bar that the angle BDC is greater than the angle CEB; much more therefore is the angle BDC greater than the angle BAC. Therefore, if from the ends of the side, &c. Q. E.D.

PROPOSITION XXII. PROBLEM.

To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third. Let A, B, C be the three given straight lines,

of which any two whatever are greater than the third, (1. 20.) namely, A and B greater than C;

A and C greater than B;

and B and C greater than A.

It is required to make a triangle of which the sides shall be equal to A, B, C, each to each.

Take a straight line DE terminated at the point D, but unlimited towards E,

make DF equal to A, FG equal to B, and GH equal to C; (1. 3.) from the center F, at the distance FD, describe the circle DKL; (post 3.)

from the center G, at the distance GH, describe the circle HLK; from K where the circles cut each other, draw KF, KG to the points F, G;

Then the triangle KFG shall have its sides equal to the three straight lines A, B, C.

Because the point Fis the center of the circle DKL,
therefore FD is equal to FK; (def. 15.)

but FD is equal to the straight line A;
therefore FK is equal to 4.

Again, because G is the center of the circle HKL;
therefore GH is equal to GK, (def. 15.)
but GH is equal to C;

therefore also GK is equal to C; (ax. 1.)
and FG is equal to B;