then every angle is congruent to its corresponding and to its alternate angle, and is supplemental to the angle on the same side of the transversal which is interior or exterior according as the first is interior or exterior.

Ex. 17. If two interior or two exterior angles on the same side of the transversal are supplemental, the straights are parallel.

Ex. 18. Two straights perpendicular to the same straight are parallel.

Ex. 19. Construct a right angle.

Ex. 20. On the ray from the vertex of a triangle costraight with a side take a sect congruent to that side. The two new end-points determine a straight parallel to the triangle's third side.

Ex. 21. On one side of any with vertex A take any two sects AB, AC and on the other side take congruent to these AB', AC'. Prove that BC' and B'C intersect, say at D. Prove BC'B'C, ABCD = ▲ B'C'D, BAD = B'AD.

Ex. 22. From two given points on the same side of a given st' find st's crossing on that given st', and making congruent's with it.

Ex. 23. Construct a triangle, given the base, an angle at the base, and the sum of the other two sides [▲ from a, b, a+c].

Ex. 24. If the pairs of sides of a quadrilateral not consecutive are congruent, they are .

Ex. 25. On a given sect as base construct an isosceles ▲. Ex. 26. If on the sides AB, BC, CA of an equilateral ▲, AD=BE=CF, then ▲ DEF is equilateral, as is ▲ made by AE, BF, CD.

CHAPTER IV.

PARALLELS.

IV. Assumption of Parallels (Euclid's Postulate).

IV. Through a given point there is not more than one parallel to a given straight.

71. The introduction of this assumption greatly simplifies the foundation and facilitates the construction of geometry.

72. Theorem. Two straights parallel to a third are parallel.

Proof. Were 1 and 2 not parallel, then there would be through their intersection point two parallels to 3, which is in contradiction to IV.

73. Theorem. If a transversal cuts two parallels, the alternate angles are congruent.

Proof. Were say BAD not = ABC, then we could through A (by III 4) take a straight making BAD' = 4 ABC [D' and

D on same side of AB], and so we would have (by 69) through A two parallels to a, in contradiction to IV.

74. Corollary to 73. A perpendicular to one of two

D

B

A

a'

FIG. 26.

parallels is perpendicular to the other also.

α

75. Theorem. If a transversal cuts two parallels, the corresponding angles are congruent.

Proof. The angle vertical to one is alternate to the other.

Ex. 27. A straight meeting one of two parallels meets the other also.

Ex. 28. A straight cutting two parallels makes conjugate angles supplemental.

Ex. 29. If alternate or corresponding angles are unequal or if conjugate angles are not supplemental, then the straights meet. On which side of the transversal?

76. Theorem. A perpendicular to one of two parallels is parallel to a perpendicular to the other. Proof. Either of the two given parallels makes (by 74) right angles with both perpendiculars, which therefore are parallel by 69.

77. Corollary to 76. Two straights respectively perpendicular to two intersecting straights cannot be parallel.

Proof. For if they were parallel, then (by 76) the intersecting straights would also be parallel.

78. Convention. When two angles are set off from the vertex of a third against its sides so that no point is interior to two, if the two sides not common are costraight, the three angles are said together to form two right angles.

79. The angles of a triangle together form two right angles.

Proof. Take alternate CBF = C and ABD =XA; then (by 69) can neither BF nor BD cut AC. A By the parallel postulate IV, then is FBD a straight.

80. Theorem. If two angles of one triangle are congruent to two of another, then the third angles

the adjacent angles ACB = × A'C'B'.

81. Theorem. Two triangles are congruent if they have a side, an adjoining and the opposite angle respectively congruent.

Proof. By 80 and 44.

Ex. 30. Every triangle has at least two acute angles. Ex. 31. If the rays of one angle are parallel or perpendicular to those of another, the angles are congruent or supplemental.

Ex. 32. In a right-angled triangle [a triangle one of whose angles is a right angle] the two acute angles are complemental (calling two angles complements which together form a right angle).

82. Theorem. In any sect AB there is always one and only one point C such that AC = BC.

Proof. Take any angle BAD at A against AB, and the angle congruent to it at B against BA and on the opposite side of a in the plane BAD; and take any sect AD on the free ray from A, and one

BF congruent to it on the free ray from B. The sect DF must cut a, say in C, since D and F are on opposite sides of a. Moreover, C is between A and B. Otherwise one of them, say A, would be between B and C. But then DA would have a point A on

D

α

B

FIG. 29.

BC, a side of triangle FBC, and so (by II 4) must meet another side. But this is impossible, since it meets FC produced at D and is parallel to BF. Since thus A= 4B, and ACD = 4 BCF [vertical], therefore (by 81) ▲ ACD = ^ BCF.

Therefore AC=BC.

If we suppose a second such point C', then on ray DC' take C'F' = DC'. Therefore (by 43) 4C'BF' = DAC = ABF, and BF' = AD=BF. Therefore F' is F and C' is C.

83. Convention. The point C of the sect AB such that ACBC may be called the bisection-point of AB, and to bisect AB shall mean to take this point C.

Ex. 33. Parallels through the end-points of a sect intercept congruent sects on any straight through its bisection-point.

Ex. 34. In a right-angled triangle the bisection-point of the hypothenuse (the side opposite the r't) makes equal sects with the three vertices.

Hint. Take one acute in the r't .