Hence, if the lines OB, O'D, intersected in I, the lines OA, O'C, now coinciding with them in their reversed position, would have the same point of intersection in I'.

Cor. 1. Conversely, two parallel lines intersected by a (80) third, make the alternate angles equal.

=

For, if the lines AB, CD, being parallel, do not make [fig. 8.] the angles AOO', OO'D equal, draw A'OB' (the student will draw the line) making / A'OO' 00'D, then will (79) A'B' be parallel to CD, but, by hypothesis, AB is parallel to CD, ... through the same point O two lines AB, A'B', have been drawn parallel to the same line CD, which is contrary to (71).

Cor. 2. The equality of two alternate angles determines: (81) 1o. The equality of all the other alternate angles, whether internal or external. (Where is this shown?)

2o. The equality of the external to the opposite internal on the same side of the secant line. (Why?)

3o. The sum of the internal angles, or the sum of the external, on the same side of the secant line, to be equal to two right angles. For, if the angle DO'E = AOF, adding / BOF to both sides, we have

Cor. 3. If the secant line be perpendicular to one of two (82) parallel lines, it will be perpendicular to the other also.

Cor. 4. Two lines parallel to the same line, are parallel to each other. [How can this be shown from (82)?]

(83)

Cor. 5. Two angles having their sides parallel, and lying (84) in the same direction, are equal. For, producing the sides of the angles a, b, till they meet, forming the angle c, we have (81, 2°), a = c = b.

Fig. 82.

Cor. 6. The opposite angles of a Parallelogram are equal. It is scarcely necessary to remark that a parallelogram is a quadrilateral, or four-sided figure, having its opposite sides parallel.

Application. Draw parallel lines upon paper with the straightedge and rightangle, and construct them in the field by aid of the cross.

(85)

%

Fig. 83.

Fig. 8.

PROPOSITION IV.

The sum of the external angles of a Polygon, formed by (86) producing the sides outward, is equal to four right angles.

Let a, b, c, d, e, be the external angles of a polygon; through any point, either within or out of the polygon, draw lines parallel to the sides of the polygon, forming the angles a', b', c', d', e', corresponding severally to the angles a, b, c, d, e; then (84) we have

▲ a = a', b = b', c = c', d = d', e = e' ; .. adding, a+b+c+d+e=a' + b' + c'+d+e+. If we denote the corresponding internal angles by A, B, C,..., of which we will suppose there are n, we shall have (74),

-=

A+ a +, B+b=+, C+c=+, ... [n]
A+B+C+...+a+b+c+... = n(+),
a+b+c+... = 2();

but (86)

a

b

Fig. 9.
Q. E. D.

a

Fig. 92.

.. subtracting, A+B+C+... [n] = (n − 2) (+)... Cor. 1. The sum of the angles of a polygon is equal to (87) two right angles taken as many times as there are sides less two. Making successively n = 6, 5, 4, 3, we have

(n − 2) (+) = 4 (+), 3 (+), 2 (+), (+); therefore Cor. 2. The sum of the angles of a Hexagon is equal to (88) eight right angles.

Cor. 3. The sum of the angles of a Pentagon is equal to (89) the sum of six right angles.

Cor. 4. The sum of the angles of a Quadrilateral is equiv- (90) alent to four right angles...

Cor. 5. All the angles of a quadrilateral may be right an- (91) gles; such a figure is called a Rectangle.

Cor. 6. The sum of the angles of a Triangle is equal to (92) two right angles.

Cor. 7. The sum of the acute angles of a Right angled (93) Triangle is equal to a right angle

Cor. 8. The external angle formed by producing one of (94) the sides of a triangle, is equal to the sum of the two opposite internal angles. Let a, b, c, be the three angles of any triangle, and c adjacent to the external angle d; we have

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Fig. 93.

Scholium. The angles of a triangle will be determined: 1o. When the three angles are equal; (how?)

2o. When two angles are equal and the third known; (how?) 3o. When two angles are given. (How?)

71

What is a hexagon? pentagon? quadrilateral? triangle? right angled triangle ?

EXERCISES.

10. By aid of the straightedge and rightangle, construct about the vertex of a triangle angles equal to those at the base, and do this for triangles of different forms.

2o. Through the vertex of any triangle draw a line paral- Fig. 10. lel to the base and prove (92).

3°. The same by drawing a parallel to one of the oblique sides. 4°. The same by drawing perpendiculars to the base.

5°. From any angle of a polygon, draw diagonals to all the other angles and prove (87).

6. From any point within a polygon, draw lines to the several angles and prove the same.

7°. Let fall a perpendicular from the right angle of a right angled triangle upon the hypothenuse, and prove that the three triangles thus formed are equiangular.

8°. Having drawn the sides of two angles respectively perpendicular to each other, prove that the angles are either equal or that one is what the other wants of two right angles.

9°. Prove that two triangles are mutually equiangular when the sides of the one are severally perpendicular to the sides of the other.

Fig. 11.

к

Fig. 12.

D

Fig. 13.

SECTION SECOND.

Equal Polygons.-First Relations of Lines and Angles.

PROPOSITION I.

If, in two Polygons, excepting three parts which are (95) adjacent, viz., two angles and an included side, or two sides and an included angle, the remaining parts, taken in the same order, are severally equal, the Polygons will be equal throughout, and the excepted parts will be equal, each to each.

=

=

B'C',

A4

B

Fig. 14.

D'

First. Let AB = A'B', B B', BC = < C = C', CD =C'D': then, applying the figure ABCD to A'B'C'D', let the point A fall on A' and the line AB take the direction A'B'; the point B will fall on B', since AB A'B' by hypothesis, and BC will take the direction B'C', because the angle BB', by hypothesis, and the point C will fall on C'; (why?) and finally, the line CD will coincide with C'D', (why?) so that the points A and D coinciding with A' and D', the line AD will coincide with A'D', and the polygons will coincide throughout, and < A will = A', AD = A'D', D = D'.

Second. Let A = A', AB = A'B', B = B', BC= B'C', C = C'; then by a superposition altogether similar to that above, it may be made to appear that the polygons will coincide throughout, and that the three adjacent parts, CD, D, DA, will be severally equal to the three, C'D', ▲ D', D'A'; and it is obvious that the same reasoning may be extended to polygons of any number of sides. Q. E. D.

Cor. 1. If two triangles have two sides and the included (96) angle of the one, equal to the two sides and included angle of the other, each to each, the triangles will be equal, and the remaining three parts of the one respectively equal to the remaining three parts of the other.

[The student should letter the figure and explain.] Fig.142

Cor. 2. If two triangles have two angles and the included (97) side of the one, severally equal to the two angles and the included

side of the other, the triangles will be equal throughout. [By what figure, and how, may this be illustrated?]

Cor. 3. The diagonal divides the parallelogram into two (98) equal triangles; ..

1°. The opposite sides of a parallelogram are equal.

2o. Parallels are everywhere equally distant.

Fig. 143.

[Letter the figure and explain carefully. What condition must be imposed upon the parallelogram in order to illustrate the last part of the corollary ?]

Cor. 4. The diagonals of a parallelogram mutually bisect, (99) or divide each other into equal parts (98, 1°), (80),

(97).

Fig. 144.

Cor. 5. An Isosceles triangle is bisected by the line (100) which bisects the angle embraced by the equal sides.'"

Cor. 6. In an isosceles triangle, the angles opposite the (101) equal sides are themselves equal; as A

=

B, and

Cor. 7. The line bisecting the vertical angle of an isos- (102) celes triangle bisects the base, to which it is also at right angles or perpendicular.

Cor. 8. Equilateral triangles are equiangular (100).

PROPOSITION II.

(103)

In any triangle, that angle is the greater which is op- (104)

posite the greater side.

Let CB be > CA; then will ▲ CAB > CBA.

For, taking CD, a part of CB, equal to CA and join

ing AD, we have

< CAB > CAD = CDA = ABD + BAD > ABD.

Q. E. D.

[Give the reasons, Consult (?), (101). (94), (?).

Fig. 15.

Cor. 1. Conversely, the side opposite the greater angle, (105) is the greater. For if the side opposite the greater angle be not the greater, it must be equal or less; it cannot be equal, for then would the angles be equal (?), neither can it be less, for then would the opposite angle be the less (104), which is contrary to the hypothesis.