To find the Ratio of two Lines.

155. Problem. To find a common measure of two given straight lines, AB, CD (fig. 87), in order to express their ratio in numbers.

Solution. a. The method of finding the common divisor is the same as that given in arithmetic for two numbers. Apply the smaller CD to the greater AB, as many times as it will admit of; for example, twice with a remainder BE.

Apply the remainder BE to the line CD, as many times as it will admit of; twice, for example, with a remainder DF.

Apply the second remainder DF to the first BE, as many times as it will admit of; once, for example, with a remainder BG.

Apply the third remainder BG to the second DF, as many times as it will admit of.

Proceed thus till a remainder arises, which is exactly contained a certain number of times in the preceding.

This last remainder is a common measure of the two proposed lines; and, by regarding it as unity, the values of the preceding remainders are easily found, and, at length, those of the proposed lines from which their ratio in numbers is deduced.

If, for example, we find that GB is contained exactly three times in FD, GB is a common measure of the two proposed lines.

EB = 1. FD + GB = 3 + 1 = 4,

CD 2. EB + FD=8+ 3 = 11,

=

AB2, CD + EB = 22+4=26;

To divide a Line into equal Parts.

consequently, the ratio of the lines AB, CD is as 26 to 11; that is, AB is 26 of CD, and CD is 1⁄2 of AB.

156. Corollary. By a like process, may be found the ratio of any two quantities, which can be successively applied to each other, like straight lines, as, for instance, two arcs or two angles.

CHAPTER X.

PROPORTIONAL LINES.

157. Theorem. If lines a a', bb', cc', &c. (fig. 88), are drawn through two sides AB, AC of a triangle ABC, parallel to the third side BC, so as to divide one of these sides AB into equal parts Aa, ab, &c., the other side AC is also divided into equal parts Aa', a' b', &c.

Demonstration. Through the points a', b', c', &c. draw the lines a'm, b'n, c'o, &c. parallel to AB.

The triangles Aaa', a'm b', b'n c', &c. are equal, by art. 53; for the sides a'm, b'n, c'o, &c. are, by art. 78, respectively equal to ab, bc, cd, &c., and are therefore equal to each other and to Aa; moreover, the angles a A a', ma'b', nb'c', &c. are equal, by art. 29, and likewise the angles Aaa', a'm b', b'nc', &c. Consequently, the sides A a', a'b', b'c', &c. are equal.

158. Problem. To divide a given straight line AB (fig. 89) into any number of equal parts.

A Line drawn Parallel to a Side of a Triangle.

Solution. Suppose the number of parts is, for example, six. Draw the indefinite line 40; 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 into six equal parts.

Demonstration. 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.

159. 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.

Demonstration. Suppose, for example, the ratio of AD: AB to be as 4 to 7. AB may then be divided into 7 equal D parts Aa, ab, bc, &c., of which AB 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.

160. Corollary. In the same way

and

AD: BD

AE: EC.

BD: AB = EC: AC.

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

Division of a Line into Pa ts p. oportional to given Lines.

162. Theorem. Conversely, if a line DE (fig. 90) is drawn so as to divide two sides AB, AC of a triangle proportionally, this line is parallel to the third side BC.

Demonstration. For the line 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.

163. 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.

Demonstration. For, by art. 160,

AE: EB AC: CD = m : n.

164. 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.

Demonstration. 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 art. 159, that

To find a Fourth proportional to three given Lines.

AC: CD'

AC: CD=m: n.

and that

CD: D'B

CD: DE = n: 0;

or, as they may be written for brevity,

AC: CD: 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 ton; 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 BC, and AE is the required line.

Demonstration. For, by art. 159,

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

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

166. 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.

Demonstration. Produce BA to E, making AE equal to Join CE.

AC.

Then the angles ACE and AEC are equal, by art. 55; and the exterior angle CAB of the triangle ACE is