To divide one side of a Triangle into parts proportional to other Sides.

equal to ACE + AEC, or to 2 CEA, and, as DAB is half of BAC, we have

DAB (2 CEA) = CEA;

ånd, therefore, by art. 31, AD is parallel to CE, and, by art. 160,

BD: DC=BA : AE,

or, since AE= AC,

BD: DC=BA : AC.

167. Problem. Through a given point P (fig. 95) in a given angle A, to draw a line so that the parts intercepted between the point and the sides of the angle may be in a given ratio.

Solution. Draw PD parallel to AB. Take DC in the same ratio to AD as the parts of the required line. Through C and P draw CPE, and this is the required line.

Demonstration. For, by art. 160,

168. Corollary. When DC is taken equal to AP, PC is equal to PE.

5

Similar Polygons and Arcs. Altitude.

CHAPTER XI.

SIMILAR POLYGONS.

169. Definition. Two polygons are similar, which are equiangular with respect to each other, and have their homologous sides proportional.

170. Definition. In different circles, similar arcs are such as correspond to equal angles at the centre. Thus the arcs AD, A'D', &c. (fig. 46) are similar.

171. Definitions. The altitude of a parallelogram is the perpendicular, which measures the distance between its opposite sides considered as bases.

The altitude of a triangle is the perpendicular, as AD (fig. 96), which measures the distance of any one of its vertices, as A, from the opposite side BC taken as a base.

The altitude of a trapezoid is the perpendicular EF (fig. 97) drawn between its two parallel sides.

172. Theorem. Two triangles ABC, DEF (fig. 98), which are equiangular with respect to each other, are similar.

Demonstration. Place the angle D upon its equal A; E must fall upon E', and F upon F; and FE' is parallel to to BC, because the angles AEF and ACB are equal. Hence, by art. 159,

AE: AC AF' : AB,

Equiangular Triangles are similar. Cases of similar Triangles.

that is,

DE: AC DF: AB.

In the same way, it may be proved that

173. Corollary. Hence, and from art. 66, it follows that two triangles are similar, when they have two angles of the one respectively equal to two angles of the other.

174. Corollary. Two right triangles are similar, when they have an acute angle of the one equal to an acute angle of the other.

175. Theorem. Two triangles are similar, when they have the sides of the one respectively parallel to those of the other.

Demonstration. If the side AB (fig. 98) is parallel to DF, and AC to DE, the angle A is, by art. 29, equal to D; and, if, moreover, BC is parallel to EF, the angles B and Fare equal, and the angles C and E are equal; so that the triangles ABC and DEF are equiangular, and therefore similar.

176. Corollary. The parallel sides are homologous.

177. Theorem. Two triangles are similar, when the sides of the one are equally inclined to those of the other, each to each, as ABC, DEF (fig. 99).

Demonstration. For if one of the triangles is turned around, by a quantity equal to the angle made by the sides of the one with those of the other, the sides of the two triangles become respectively parallel, and they are, therefore, by art. 175, equiangular and similar.

Cases of similar Triangles.

178. Corollary. Two triangles are similar, when the sides of the one are respectively perpendicular to those of the other, and the perpendicular sides are homologous.

179. Theorem. Two triangles ABC, DEF (fig. 98) are similar, if they have an angle A of the one equal to an angle D of the other, and the sides including these angles proportional, that is,

AB: DF = AC: DE.

Demonstration. Place the angle D upon A; E falls upon E', and Fupon F; and EF is parallel to BC, by art. 162, because

=

BAFE

Hence, by art. 30, the angle C AEFE,

and

that is, the triangles ABC and DEF are equiangular, and, by art. 172, similar.

180. Theorem. Two triangles ABC, DEF (fig. 98) are similar, if they have their homologous sides proportional, that is,

AB: DE AC: DE = BC: EF.

Demonstration. Take AE' =

DE, and draw E'F' par

allel to BC. The triangles AE'F' and ABC are similar, by art. 175, and we are to prove that AE'F' is equal to DEF.

Now, by art. 159,

AE AC = AF': AB,

and, by hypothesis,

DE or AE: AC: = DF: AB.

Right Triangle divided into two similar Right Triangles.

Hence, on account of the common ratio AE : AC,

AF' : AB = DF : AB ;

that is, AF and DF are in the same ratio to AB, and are consequently equal.

In the same way it may be proved that E'F' and EF, being in the same ratio to BC, are equal; and as the triangle DFE has its sides equal to those of AE'F", it is equal to AE'F', and is, therefore, similar to ABC.

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181. Theorem. Lines AF, AG, &c. (fig. 100), drawn at pleasure through the vertex of a triangle, divide proportionally the base BC and its parallel DE, so

that

DI: BF IK: FG KL: GH, &c. Demonstration. Since DI is parallel to BF, the triangles ADI, ABF are equiangular, and give the proportion, DI: BF = AI : AF;

also, since IK is parallel to FG,

AI: AF = IK : FG ;

and, therefore, on account of the common ratio AI: AF

It may be shown in like way, that

IK : FG = KL : GH, &c.

182. Corollary. When BC is divided into equal parts, the parallel DE is likewise divided into equal parts.

183 Theorem. The perpendicular AD (fig. 101) upon the hypothenuse BC of the right triangle BAC from the vertex A of the right angle, divides the triangle into two triangles BAD, CAD, which are similar to each other and to the whole triangle BAC.