THEOREM III.

If a straight line intersect two parallels, the corresponding inner and outer angles will be equal to each other; also the alternate angles.

Let the straight line EF intersect the two parallel straight lines AB, CD, in AG and H. Then will any two corresponding inner and outer angles, as GHD and EGB, be equal.

C

E

m

n

-B

H

D

F

For, since EF is a straight line, its part HG has the same direction from H that its part GE has from G (Def. 6, Sec. I); and since HD and GB are by hypothesis parallel, HD has the same direction from H that GB has from G (Def. 4, Sec. IV). Therefore, the difference in direction between HG and HD is equal to the difference in direction between GE and GB; that is, the angle GHD is equal to the angle EGB (Def. 1, Sec. IV).

Again, any two alternate angles, as AGH and GHD, are equal to each other. For AGH is equal to its opposite angle EGB (Theo. II). But EGB has just been proved equal to GHD. Consequently, AGH is equal to GHD (Ax. 3).

Therefore, if a straight line intersect, etc.

Cor. 1. It is evident that if AB is not parallel to CD, but takes some other direction, as mGn, the corresponding inner and outer angles will be unequal; also, the alternate angles.

Cor. 2. If two angles have their sides parallel,

each to each, and directed the same way from the vertex, they are equal.

THEOREM IV.

If a straight line intersect two parallels, the sum of the two inner angles on the same side will be equal to two right angles.

Let EF intersect the two parallels AB, CD, in G and Then will BGH and

H.

GHD be together equal to

two right angles.

For the sum of the angles GHC and GHD is two

A

F

E

B

G

right angles (Theo. I). But GHC is equal to its alternate angle BGH (Theo. III). Therefore, the sum of BGH and GHD is equal to two right angles. Hence, if a straight line, etc.

Cor. 1. If BGH is a right angle, GHD must also be a right angle. Hence, if a straight line is perpendicular to one of two parallels, it is perpendicular to the other.

Cor. 2. If two straight lines are both perpendicular to a third, they are parallel.

EXERCISES..

1. Prove that the sum of all the adjacent angles made by any number of straight lines meeting in one point is equal to four right angles.

2. Prove that if one of the four angles made by Evans' Geometry.-2

two straight lines intersecting each other be a right angle, each of the others will be a right angle.

3. Prove that the alternate outer angles EGA and FHD (Figure to Theo. III) are equal to each other.

4. Prove that the sum of the two outer angles on the same side, EGB and FHD (Fig. Theo. IV), is equal to two right angles.

5. Point out all the angles equal to EGB (Fig. Theo. IV); also, all the angles equal to EGA; and show in each case why they are equal.

I

SEC. V.-TRIANGLES.

DEFINITIONS.

1. A PLANE FIGURE is a portion of a plane bounded on all sides by lines.

2. A POLYGON is a plane figure bounded by straight lines.

3. A TRIANGLE is a polygon of three sides.

If it has one right angle it is called a right-angled triangle. The side opposite the right angle is called the HYPOTENUSE.

A triangle which has no right angle is called an oblique-angled triangle.

An isosceles triangle is one which has two equal sides.

4. Any side of a triangle may be considered as its base, and the opposite angle as its vertex; but in an isosceles triangle, that side is usually called the base which is not equal to either of the others.

The ALTITUDE of a triangle is the perpendicular let fall from the vertex on the base, or the base produced.

5. A triangle, or other polygon, is called equiangular when all its angles are equal; equilateral, when all its sides are equal.

6. Two polygons are called mutually equilateral or mutually equiangular, if the sides or angles of the one are equal to the sides or angles of the other, each to each, taken in the same order.

THEOREM V.

The three interior angles of any triangle are together equal to two right angles.

Let ABC be any triangle. It is to be proved. that the angles ABC, BCA, CAB, are together equal A4 to two right angles.

E

D

Produce AB to any point D, and draw BE parallel to AC. Now, the sum of the three angles ABC, CBE, EBD, is equal to two right angles (Cor. 1, Theo. I). But because CB intersects the parallels AC and BE, the angle CBE is equal to its alternate angle BCA (Theo. III); also, because AD intersects the same parallels, the angle EBD is equal to its corresponding inner angle CAB. Therefore, the sum of ABC, BCA, CAB, is equal to two right angles.

That is, the three interior angles of a triangle, etc.

Cor. 1. If one side of a triangle be produced, the exterior angle will be equal to the sum of the two opposite interior angles. For the sum of CBE and EBD, that is, the whole exterior angle CBD, is equal to the sum of BCA and CAB.

Cor. 2. A triangle can not have more than one right angle; for, if it had two, the third angle would be nothing. Still less can a triangle have more than one obtuse angle.

THEOREM VI.

If two straight lines be drawn from the extremities of one side of a triangle to a point within, their sum will be less than that of the other two sides of the triangle.

From the extremities of AB, let straight lines be drawn to a point D within the triangle ABC. It is

to be proved that the sum of AD and DB is less than the sum of AC and CB.

E

B

Produce AD to meet BC in E. Now, since AE is a straight line, it is less than the sum of AC and CE, which form a broken line (Ax. 10). Therefore, if we add EB to both, it is evident that the sum of AE and EB is less than the sum of AC, CE, and EB, that is, less than the sum of AC and CB (Ax. 6). In the same manner it may be shown that the sum of AD and DB is less than the sum of AE and EB; still more, then, we may conclude it is less than the sum of AC and CB.

Therefore, if two straight lines, etc.