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Proof of (a), (b), (c).—This series of propositions is proved by means of Theorem VII. When the straight lines meet, they form a triangle, of which one of the alternate angles is the exterior angle and the other an interior angle; therefore these angles are not equal: consequently, when the alternate angles are equal, the straight lines cannot meet.

Proof of (a′), (b′), (c′).—To prove the second series of propositions, we have to receive as self-evident the fact that two straight lines which intersect cannot both be parallel to the same straight line.

Sum of the Angles of a Triangle.-By means of the foregoing pro perties of parallels we are able to prove the following remarkable and important theorem, namely, that, whatever be the form of a triangle, all its angles are together equal to two right angles. We may show this materially as follows:-Cut out of paper a triangle, A B C (Fig. 68a). Fold the paper so that a part, DC, of the line B C falls on the other

A 4 4

FIG. 68a.

FIG. 686.

A
BC

FIG. 68c.

part, DB, and the line formed by the crease passes through A (Fig. 686). By this means we form two right angles at D. Fold the triangle again, by bringing all the three angular points to the point D. It will then be found that the three angles of the triangle exactly cover the two right angles.

This can be shown, also, by drawing three angles equal to those of a triangle, in such a manner that they shall be seen to coincide with two right angles. Thus we may suppose a line drawn through A parallel

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to BC (Fig. 69a), or we may continue B C, and draw from C a line parallel to B A (Fig. 696).

Deductions from the preceding Theorem.-From the proposition that the three angles of a triangle are always together equal to two right angles, we may make several simple deductions.

(a). If two triangles have two angles of the one respectively equal to

two angles of the other, the third angle of the one is equal to the third angle of the other.

Hence when two triangles have two angles of the one respectively equai to two angles of the other and a side opposite an angle in one equal to the side opposite the equal angle in the other, the triangles are equal in all respects.

For the third angle of the one is equal to the third angle of the other, and therefore the triangles are equal by Theorem VI.

(b). If in a triangle an angle is a right angle, each of the other two is less than a right angle.

(c). If one angle of a triangle be obtuse, each of the others must

be acute.

This theorem, together with those forming the exercise on page 43, enables us to establish some properties of lines having many useful applications, although they are not essential links in the general chain of geometrical reasoning.

1. Every oblique line drawn from a point to a straight line is greater than the perpendicular from the same point to the straight line. For if CD be the perpendicular, and C F be an oblique line from C to A B, then ACDF has a right angle, C D F, and, therefore, an acute angle, CFD. Consequently CF is greater than CD.

2. Two oblique lines, the feet of which are equidistant from the foot of a perpendicular let fall from the same point, are equal to one

another.

C

A

E

D

F

B

FIG. 700.

If CE and CF be the equal obliques, and CD the perpendicular (Fig. 70a), the triangles AED AFD are equal by Theorem V.; therefore CE = CF.

Definition.-The distance of the foot of the oblique from the foot of the perpendicular is termed the departure of the oblique.

3. Of two oblique lines from the same point, that which has the greater departure is the greater; and, conversely, the greater has the greater departure.

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Let CE and C F be two oblique lines, and CD the perpendicular (Fig. 70b); erect a perpendicular G H at G, the middle of the line EF; join HE, then HEHF. Now the straight line CE is shorter than CH+HE, or than its B equal C F.

Conversely the longer cblique has the

greater departure. Let CF be greater than CE, then F is farther from the perpendicular than E; for, if it were nearer, CE would be less than CF; if at the same distance, then CE and CF would be equal to one another; therefore F is farther from the perpendicular than E.

Arithmetical Exercises.

1. In a right-angled triangle one acute angle is double of the other : give the magnitude of the angles.

2. In a triangle two angles are equal and the third is double either of the first find the angles.

3. Three angles of a four-sided figure are respectively 27°, 89°, and III: find the fourth.

4. One angle of a parallelogram is 25°: find the others.

5. Two angles of a triangle are 25° 13′ 15′′ and 56° 14′ 13′′: find the third.

6. Two angles of a triangle are 57° 13′ 45′′ and 95° 46′ 15′′: find the three exterior angles.

7. In an isosceles trapezoid one angle is 39° 58′ 6": find the others. 8. The angle at the vertex of an isosceles triangle is 25° 15′ 46′′ : find the angle at the base.

9. In a five-sided figure the angles are equal: find them.

10. What is the magnitude of the angle in a six-sided figure, the angles of which are equal?

II. The same for a seven-sided figure.

12. The same for an eight-sided figure.

13. In an isosceles triangle each angle at the base is double the vertical angle: find the number of degrees in the angles.

THEOREMS ON CHAPTER V.

PARALLELS.

DEFINITION.

35. Two straight lines in the same plane which do not meet, however far they may be produced, are parallel. Axiom. Two straight lines passing through the same point cannot both be parallel to the same straight line.

THEOREM XVI.

If a straight line, cutting two other straight lines, makes either a pair of alternate angles equal or a pair of corresponding angles equal, or the two interior angles on one side of the cutting line equal to two right angles, these two straight lines shall be parallel.

E

Let EF cut A B and C D in G and H, making the alternate angles AGH, GHD equal to one another, then shall A B and CD be parallel.

A

H

C

B

For if A B and C D were to meet, as, for instance, in a point M, then G H M would be a triangle, and the exterior angle A G H would be greater than the interior angle G H M.

But A G H G H M, and therefore A B and C D do

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not meet, that is to say, they are parallel.

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Again, when the corresponding angles E G B, GHD are equal, since EGB AG H, the alternate angles AGH, G H D are also equal, and consequently A B and CD are parallel.

And when the two interior angles B G H and GHD are together equal to two right angles, since B G H and A G H are equal to two right angles, the alternate angles AGH, GHD are equal, and consequently A B and C D are parallel.

THEOREM XVII.

When a straight line cuts two parallel straight lines, it makes the alternate angles equal, the corresponding angles equal, and the interior angles on the same side equal to two right angles.

Let the straight lines EF cut the parallel straight lines A B, CD in G and H, then shall ▲ AGH = ▲ G HD, ▲ E G B = ▲ GHD,

And BGH, GHD be together equal to two right angles.

For if A G H be not equal to GHD, suppose M G to be the line

E

M

F

H

B

making with GH, ▲ MGH = 4G HD. Then, since the alternate angles M G H, GHD are equal, M G and C D are parallel, and through the same point G there are two straight lines M G and AG both parallel to CD, which is impossible.

Therefore A G H, G H D must be equal to one another. Since A G H = EG B, therefore also G H D = EG B. Since AGH, BGH are together equal to two right angles, therefore G H D and B G H are together equal to two right angles.

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