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If a straight line EF cut the two straight lines AB, CD, it makes with either of them four angles which we will denote by (1), (2), (3), (4); (I), (II), (III), (IV).

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Then (2) = (II), for these are the supplements of (1) and (I).

Also (3) and (4) = (III) and (IV) each to each, being vertically opposite to the angles (1), (2), (I), (II).

Also (3) = (I), for each is equal to (1).

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The angles (1) and (I) are called exterior and interior angles.

The angles (3) and (I) are called alternate angles.

The angles (2) and (I) are called interior angles on the same side of EF.

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Hence HC is situated with regard to GA precisely as GB with regard to HD.

Therefore if GB, HD being produced meet towards B and D, HC, GA being produced must meet towards C and A, In which case two straight lines would enclose a space, which is absurd.

Therefore AB, CD do not meet either towards B and D, or towards C and A.

COR. 1. If either of the angles at G be equal to the corresponding angle at H, AB must be parallel to CD.

COR. 2. If the alternate angle AGH is equal to the alternate angle GHD, AB is parallel to CD.

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COR. 3. If the two interior angles BGH, GHD are together equal to two right angles, AB is parallel to CD.

For BGH+ GHD will then be equal to BGH+ EGB, and therefore EGB= GHD.

THEOREM VIII.

If AB be parallel to CD, then shall the angle EGB be

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For if not let EGM be equal to GHD. Then GM is parallel to HD.

Thus through the same point G two different lines are drawn each parallel to CD, which is absurd.

COR. 1. The other angles at G are equal to the corresponding angles at H.

COR. 2. The alternate angles AGH, GHD are equal.

COR. 3. The two interior angles BGH, GHD are together equal to two right angles.

THEOREM IX.

Straight lines AB, CD parallel to the same line KL are

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If one side BC of a triangle ABC be produced, the ex

terior angle ACD is equal to the two interior and opposite angles A and B; and the three

angles A, B, and C are equal to two right angles.

Through C draw CE pa

rallel to BA.

B

E

Then the alternate angle ACE = the alternate angle A,

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and the exterior angle ECD the interior angle B.

Therefore the whole ACD = A + B.

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Again,

A+B= ACD.

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Add to each C, then

A+B+C ACD+ACB

=

=2R.

CCR. 1. If two angles of one triangle be equal to twó angles of another, the third angle of the first is equal to the third angle of the second.

COR. 2. The exterior angle is greater than either of the interior and opposite angles.

THEOREM XI.

If in any triangle ABC the side AB be equal to the side AC, then shall the angle C be equal

to the angle B.

Let AD be the perpendicular drawn from A to BC.

A

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=

COR. 1. Also BAD CAD. Hence in an isosceles tri angle the perpendicular bisects the base and the angle at the

vertex.

COR. 2. All the angles of an equilateral triangle are equal.

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