PROBLEM VII.

162. To construct a triangle of which the sides shall be equal to three given sects, but any two whatever of these sects must be greater than the third.

D

G

H

a

で

GIVEN, the three sects a, b, c, any two whatever greater than the third.

b, c.

REQUIRED, to make a triangle having its sides equal respectively to a,

CONSTRUCTION. On a line DF, by 133, take a sect OG equal to one of the given sects, as b. From O as a center, with a radius equal to one of the remaining sects, as c, describe, by 102, a circle.

From G as a center, with a radius equal to the remaining sect a, describe an arc intersecting the former circle at K. Join KO and KG. KGO will be the triangle required.

PROOF. By the construction and the equality of all radii of the same circle, the three sides GK, KO, and OG are equal respectively to a, b, c.

LIMITATION. It is necessary that any two of the sects should be greater than the third, or, what amounts to the same, the difference of any two sides less than the third. For, if a and c were together less than b, the circles in the figure would not meet; and, if they were together equal to b, the point K

would be on OG, and the triangle would become a sect. But as we have proved, in 156, that any two sides of a triangle are together greater than the third side, our solution of Problem VII will apply to any three sects that could possibly form a triangle. 163. COROLLARY. If a tri

angle be made of hinged rods, though the hinges be entirely free, the rods cannot turn upon them; the triangle is rigid.

PROBLEM VIII.

164. At a given point in a given line, to make an angle equal

to a given angle.

A A

GIVEN, the point A, and the line AB and

C.

REQUIRED, at A, with arm AB, to make an angle = C.

CONSTRUCTION. By 100, join any two points in the arms of C, thus forming a ▲ DCE. By 162, make a triangle whose three sides shall be equal to the three sides of DCE; making the sides equal to CD and CE intersect at A.

.. FAG = X DCE.

(129. Triangles with three sides respectively equal are congruent.)

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 = † PCB.

PD is || AB.

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

.. × KHB - 4 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.