THEOREM XXVI.

185. The locus of points from which perpendiculars on two given intersecting lines are equal, consists of the two bisectors of the angles between the given lines.

HYPOTHESIS. Given AB and CD intersecting at O, and P any point such that PM | AB = PN 1 CD.

(179. Right triangles having hypothenuse and one side equal are congruent.) Therefore all points from which perpendiculars on two intersecting lines are equal, lie on the bisectors of the angles between them.

186. INVERSE.

HYPOTHESIS. P, any point on bisector.

CONCLUSION. PM = PN.

PROOF. Δ ΡΟΜ Ξ PON.

(176. Triangles having two angles and a corresponding side equal in each are

congruent.)

Hence, by 110, the two bisectors of the four angles between AB and CD are the locus of a point from which perpendiculars on AB and CD are equal.

Intersection of Loci.

187. Where it is required to find points satisfying two conditions, if we leave out one condition, we may find a locus of points satisfying the other condition.

Thus, for each condition we may construct the corresponding locus; and, if these two loci have points in common, these points, and these only, satisfy both conditions.

THEOREM XXVII.

188. There is one, and only one, point from which sects to three given points not on a line are equal.

B

D

HYPOTHESIS. Given A, B, and C, not in a line.

CONSTRUCTION AND PROOF. By 183, every point to which sects from A and B are equal, lies on DD, the perpendicular bisector of AB. Again: the locus of points to which sects from B and C are equal is the perpendicular bisector FF.

Now, DD and FF' intersect, since, if they were parallel, AB, which is perpendicular to one of them, would be perpendicular to the other also, by 170, and ABC would be one line. Let them intersect in O. By 95, they cannot intersect again.

Therefore, by 183, O is on the perpendicular bisector of AC. There

fore

COROLLARY. The three perpendicular bisectors of the sides of a triangle meet in a point from which sects to the three vertices of the triangle are equal.

PROBLEM X.

190. To find points from which perpendiculars on three given lines which form a triangle are equal.

GIVEN, a, b, c, three lines intersecting in the three distinct points A, B, C.

SOLUTION. The locus of points from which perpendiculars on the two lines a and b are equal consists of the two bisectors of the angles between the lines.

The locus of points from which perpendiculars on the two lines and are equal, similarly consists of the two lines bisecting the angles between b and c. Our two loci consist thus of two pairs of lines. Each line of one pair cuts each line of the second pair in one point; so that we get four points, O, 01, 02, 03, common to the two loci.

191. By 185, the four intersection points of the bisectors of the angles between a and b, and between b and c, lie on the bisectors of the angles between a and c. Therefore

COROLLARY. The six bisectors of the interior and exterior angles of a triangle meet four times, by threes, in a point.

CHAPTER VIII.

POLYGONS.

I. Definitions.

192. A Polygon is a figure formed by a number of lines of which each is cut by the following one, and the last by the first.

193. The common points of the consecutive lines are called the Vertices of the polygon.

194. The sects between the consecutive vertices are called the Sides of the polygon.

195. The sum of the sides, a broken line, makes the Perimeter of the polygon.

196. The angles between the consecutive sides and towards the enclosed surface are called the Interior Angles of the polygon. Every polygon has as many interior angles as sides.

197. A polygon is, said to be Convex when no one of its interior angles is reflex.

198. The sects joining the vertices not consecutive are called Diagonals of the polygon.

199. When the sides of a polygon are all equal to one another, it is called Equilateral.

200. When the angles of a polygon are all equal to one another, it is called Equiangular.