For if BC be not equal to EF, one of them must be

greater. Let BC be the greater, and cut off (1.1) BG equal to EF. Then in the triangles ABG, DEF, AB, BG and ABG are equal, each to each, to DE, EF and DEF; therefore (I. 7) AGB is equal to DFE; but DFE is equal (Hyp.) to ACB; therefore

A

D

B

GC E

F

AGB is equal to ACB, an exterior angle of the triangle AGC equal to an opposite interior angle ACG, which is impossible (I. 18). Therefore BC is not unequal to EF; that is, it is equal to it, and the other parts of the triangles are equal by (I. 7).

NOTE.-Let the student write out the complete analysis.

Scholium. It will be observed, by comparing Props. (4), (7), (8), (24), that there are three cases in which triangles may be proved equal in all their parts:

1. When they have three sides of one equal to three sides of the other, each to each (I. 4);

2. When they have two sides and included angle of one equal to two sides and included angle of the other, each to each (I. 7);

3. When they have two angles and a side of one equal to two angles and a side of the other, each to each (I. 8), (I. 24).

DEFINITION.

If a line meet two parallel lines, the various angles have different

names.

1. d and e, taken in relation to each other, are called alternate, as also c and f.

fle

bla dc

2. c and e, or d and f, are interior angles on the same side. 3. a and e, or b and f, are respectively exterior and opposite interior angles on the same side.

Proposition 25.

Theorem. If a straight line falling on two straight lines, make the alternate angles equal to each other, the two straight lines will be parallel.

Let EF, falling on AB and CD, make the angles AEF, EFD equal to each other; then will AB and CD be parallel.

For if they be not parallel, they

will meet if produced.

meet towards B and D.

Let them

Then E, F

A

E

B

toward B and D. In the same way it may be proved they will not meet toward A and C. They are therefore parallel (Def. 4).

Proposition 26.

Theorem. Two straight lines will be parallel

1. When the exterior and opposite interior angle on the same side are equal to each other;

2. When the two interior angles on the same side are equal to two right angles.

1. Let EF, falling on AB and CD, make EGB equal to GHD; AB and CD will be parallel.

For EGB is equal (I. 17) to AGH; therefore AGH is equal to GHD; but they are alternate angles; therefore AB and CD are parallel (I. 25).

then AGH= GHD, and they are alternate angles; therefore

AB and CD are parallel (I. 25).

Proposition 27.

Theorem. If a straight line fall on two parallel straight lines, it makes

1. The alternate angles equal to each other.

2. The exterior angle equal to the opposite interior angle on the same side.

3. The two interior angles on the same side together equal to two right angles.

Let the straight line EF fall on the two parallel lines AB, CD.

AB and KL, have been drawn, both parallel to CD, which is impossible (Ax. 11). Hence AGH is not unequal to GHD.

2. The exterior angle EGB will be equal to the opposite interior angle GHD.

For EGB is equal (I. 17) to AGH, and AGH has been proved equal to GHD; therefore EGB is equal to GHD.

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

For EGB is equal to GHD; add to both BGH, and EGB+BGH=BGH+ GHD; but (I.15) EGB+BGH=2R;

... BGH+ GHD = 2R.

Corollary 1.-KL and CD will meet on that side of EF on which the interior angles are less than two right angles.

Let KGH, GHC be less than two right angles. If KL and CD do not meet toward K and C, they are parallel, or they meet toward Land D. They are not parallel, for then KGH and GHC would be equal to two right angles. Neither do they meet toward Land D; for then LGH and GHD would be two angles of a triangle, and less than two right angles. But if KGH, GHC are less than two right angles, LGH, GHD are greater. Hence KL and CD will meet toward K and C.

Corollary 2.-If a straight line is at right angles to one of two parallel lines, it is also at right angles to the other.

For if BGH is a right angle, GHD will also be a right angle; since they are together equal to two right angles.

Proposition 28.

Theorem. Straight lines which are parallel to the same straight line are parallel to each other.

Let AB, CD be each of them parallel to EF; AB is parallel

A

Draw GHK cutting all three. Then, because AB, EF are parallel, the alternate angles AGH, GHF are equal (I. 27) to each other; and because EF, CD are parallel, the exterior angle GHF is equal (I. 27) to the opposite interior angle HKD. Therefore A GH, HKD are equal to each other; and they are alternate; hence AB and CD are parallel (I. 25).

E

K

C

D

Proposition 29.

Problem. To draw a straight line through a given point

parallel to a given straight line.

Let A be the point, and BC the line; it is required to

draw through A a line parallel to BC.

In BC take any point D, and

E

F

A

join AD; at A make (I. 5) the angle DAE equal to the angle ADC, and produce EA to F.

B

D

C

Then, because the alternate angles
EAD, ADC are equal, EF is parallel (I. 25) to BC.

Proposition 30.

Theorem. If one side of a triangle be produced, the exterior angle is equal to the two opposite interior angles; and the three interior angles of any triangle are equal to two right angles.