A line drawn Parallel to a Side of a Triangle.

Solution. Suppose the number of parts is, for example, six. Draw the indefinite line AO; take AC of any convenient length, apply it six times to AO. Join B and the last point of division D by the line BD, draw CE parallel to DB, and AE, being applied six times to AB, divides it into six equal parts.

Proof. For if, through points of division of AD, lines are drawn parallel to DB, they must, by the preceding theorem, divide AB into six equal parts, of which AE is

one.

160. Theorem. If a line DE (fig. 90) is drawn through two sides AB, AC of a triangle ABC, parallel to the third side BC, it divides those two sides proportionally, so that we have

AD: AB= AE: AC.

Proof. a. Suppose, for example, the ratio of AD: AB to be as 4 to 7. AB may then be divided into 7 equal parts A a, ab, bc, &c., of which AD contains 4; and if lines a a', bb', cc', &c. are drawn parallel to BC, AC is divided into 7 equal parts A a', a' b', b' c', &c., of which AE contains 4. The ratio of AE to AC is, therefore, 4 to 7, the same as that of AD: AB.

161. Scholium. b. The case in which AD and AB are incommensurable, is included in this demonstration by the reasoning of § 98.

and

162. Corollary. In the same way

AD: BD = AE: EC.

BD: AB= EC : AC.

163. Theorem. Conversely, if a line DE (fig, 90)

Division of a Line into Parts proportional to given Lines.

is drawn so as to divide two sides AB, AC of a triangle proportionally, this line is parallel to the third side BC.

Proof. For the line, which is drawn through the point D parallel to BC must, by the preceding proposition, pass through the point E, so as to divide the side AC proportionally to AB, and must therefore coincide with the proposed line DE.

164. Problem. To divide a given straight line AB (fig. 91) into two parts, which shall be in a given ratio, as in that of the two lines m to n.

Solution. Draw the indefinite line AO. Take AC =m and CD=n. Join DB, through C draw CE parallel to DB; and E is the point of division required.

Proof. For, by § 161,

AE: EB AC: CD: =m: n.

165. Problem. To divide a given line AB (fig. 92) into parts proportional to any given lines, as m, n, o,

&c.

Solution. Draw the indefinite line AO. Take

AC = =m,

CD=n, DE=0, &c.

Join B to the last point E, and draw CC', DD', &c. parallel to BE. C', D', &c. are the required points of division.

Proof. For, if AE is divided into parts equal each of them to the greatest common divisor of m, n, o, &c., and if, through the points of division, lines are drawn parallel to BE; it appears, from inspection, as in § 160, that

and that

AC: C'D' AC: CD=m: n.

=

To find a Fourth proportional to three given Lines.

C'D': D'B= CD: DE=n: 0;

or, as they may be written for brevity,

AC: C'D': D'B=m:n: o.

165. Problem. To find a line, to which a given line AB (fig. 93) has a given ratio, as that of the lines m to n; in other words, to find the fourth proportional to the three lines m, n, and AB.

Solution. Draw the indefinite line AB, take

Join CB, draw DE parallel to BC, and AE is the required line.

Proof. For, by § 160,

AB: AE=AC: AD: =m: N.

166. Corollary. By making n equal to AB in the preceding solution, we find a third proportional to the two lines m and AB.

167. Problem.

To divide one side BC (fig. 94),

of a triangle ABC into two parts proportional to the other two sides.

Solution. Draw the line AD to bisect the angle BAC, and D is the required point of division, that is,

BD: DC=AB: AC.

Proof. Produce BA to E, making AE equal to AC. Join CE.

Then the angles ACE and AEC are equal, by § 55; and the exterior angle CAB of the triangle ACE is equal to ACE + AEC, or to 2 CEA, and, as DAB is half of BAC, we have

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

́and, therefore, by § 31, AD is parallel to CE, and, by & 161,

BD: DC=BA : AE,

or, since AE=AC,

BD: DC= BA: AC.

168. 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.

169. Corollary. When DC is taken equal to AD, PC is equal to PE.

CHAPTER XI.

SIMILAR POLYGONS.

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

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

Similar Polygons and Arcs. Equiangular triangles 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, as EF (fig. 97), drawn between its two parallel sides.

172. Theorem. Two triangles ABC, DEF (fig. 99), which are equiangular with respect to each other, are similar.

Proof. Place the angle D upon its equal A; E must fall upon E', and F upon F; and FE' is parallel to BC, because the angles AEF and ACB are equal. Hence, by § 160,

that is,

AE: ACAF': AB,

DE: AC DF: AB.

=

In the same way, it may be proved that

DE: AC: = EF: BC=DF: AB.

173. Corollary. Hence, and from § 67, 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