CHAPTER VI.

PARALLELS.

THEOREM XIX.

165. If two lines cut by a transversal make alternate angles equal, the lines are parallel.

This is the contranominal of 142, part of which may be stated thus: If two lines which meet are cut by a transversal, their alternate angles are unequal.

166. COROLLARY. If two lines cut by a transversal make corresponding angles equal, or interior or exterior angles on the same side of the transversal supplemental, the lines are parallel.

For we know, by 113, that either of these suppositions makes also the alternate angles equal.

EXERCISES. 21. From a given point without a given line, draw a line making an angle with the given line equal to a given angle.

PROBLEM IX.

167. Through a given point, to draw a line parallel to a given line.

CONSTRUCTION. In AB take any point, as C. By 100, join PC. At P in the line CP, by 164, make ₫ CPD

PD is | AB.

=

PCB.

PROOF. By construction, the transversal PC makes alternate angles equal; CPD = 4 PCB,

.. by 165, PD is || AB.

EXERCISES. 22. Draw, through a given point between two lines which are not parallel, a sect, which shall be terminated by the given lines, and bisected at the given point.

23. If a line bisecting the exterior angle of a triangle be parallel to the base, show that the triangle is isosceles.

24. Sects from the middle point of the hypothenuse of a right-angled triangle to the three vertices are equal.

25. Through two given points draw two lines, forming, with a given line, a triangle equiangular with a given triangle.

26. AC, BD, are equal sects, drawn from the extremities of the sect AB on opposite sides of it, such that the angles BAC, ABD, are together equal to a straight angle. Prove that AB bisects CD.

THEOREM XX.

168. If a transversal cuts two parallels, the alternate angles

HYPOTHESIS. AB || CD cut at H and K by transversal FG.
CONCLUSION. X HKD
X KHB.

=

PROOF. A line through H, making alternate angles equal, is parallel to CD, by 165.

By our assumption, 99, two different lines through H cannot both be parallel to CD;

by 31, the line through H CD is identical with the line which makes alternate angles equal.

But, by hypothesis, AB is parallel to CD through H,

X KHB = ¥ HKD.

169. COROLLARY I. If a transversal cuts two parallels, it makes the alternate angles equal; and therefore, by 113, the corresponding angles are equal, and the two interior or two exterior angles on the same side of the transversal are supplemental.

170. COROLLARY II. If a line be perpendicular to one of two parallels, it will be perpendicular to the other also.

171. CONTRANOMINAL OF 168. If alternate angles are unequal, the two lines meet.

So, if the interior angles on the same side of the transversal are not supplemental, the two lines meet; and as, by 143, they cannot meet on that side of the transversal where the two interior angles are greater than a straight angle, therefore they must meet on the side where the two interior angles are together less than a straight angle.

172. CONTRANOMINAL OF 99. Lines in the same plane parallel to the same line cannot intersect, and so are parallel to one another.

THEOREM XXI.

173. Each exterior angle of a triangle is equal to the sum of the two interior opposite angles.

HYPOTHESIS. ABC any A, with side AB produced to D.
CONCLUSION. CBD = A + XC.

=

PROOF. From B, by 167, draw BF || AC. Then, by 168, C = CBF, and by 169, 4 A = 4 DBF;

.. by adding, A + C = FBD + CBF = CBD. 4 ¥

EXERCISES. 27. Each angle of an equilateral triangle is two-thirds of a right angle. Hence show how to trisect a right angle.

28. If any of the angles of an isosceles triangle be twothirds of a right angle, the triangle must be equilateral.

THEOREM XXII.

174. The sum of the three interior angles of a triangle is equal to a straight angle.

HYPOTHESIS. ABC any ▲.

CONCLUSION. CAB+ B + C = st. X.
PROOF. Produce a side, as CA, to D.

By 173, B + C = X DAB.

Add to both sides & BAC.

.. CAB+ B + C = DAB+ BAC = st. X.

175. COROLLARY. In any right-angled triangle the two

acute angles are complemental.