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Problem 144.

To find a third proportional to two given straight lines AB and CD, when the required line is greater than either of the given lines.

1. At A in AB, draw a line AE equal to CD, and at any

angle with AB.

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2. Join BE, and produce AE and AB indefinitely.
3. From A as centre, with AB as radius, describe an arc

cutting A E produced in F.
4. Through F, draw FG parallel to EB (Pr. 9), cutting AB

produced in G. Then AG is the required third proportional to the given lines AB and CD, and greater than either of them.

Problem 145.

To divide a given straight line AB into extreme and mean proportion.

1. Bisect AB in C (Pr. 1), and from one extremity, say B,

erect a perpendicular BD equal to BC.

2. Join AD, and from the centre D, with radius DB, de3. From centre A, and radius AE, describe the arc EF.

scribe the arc BE.

Then the given line AB is divided in extreme and mean proportion in the point F.

NOTE.—"A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less ” (Euc. VI., Def. 3).

Problem 146.

To divide a given straight line AB successively into its half, third, fourth, fifth, &c.

1. On AB, construct any parallelogram ABCD, and draw

the diagonals AC, BD, intercepting each other in E.

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2. Draw line Fi parallel to AD (Pr. 9) cutting AB in :
3. Draw line C, cutting BD in 3, and draw line 3} parallel

to AD, cutting A B in .
4. Draw line C}, cutting BD in 4, and through 4 draw line

45 parallel to AD, cutting AB in , &c., &c. The divisions thus obtained are the required half, third, fourth, &c., of the given straight line AB.

Problem 147.

To divide any given straight line AB in the point C, so that AC : CB::2:3.

1. From point A, draw a straight line AD of indefinite

length, and at any angle to AB.

2. On AD, mark off any two equal distances A 1, 1 2, and

from 2, mark off three similar distances to 5 ; and join 5 B.

3. Through 2, draw 2 C parallel to 5 B (Pr. 9). Then the

given straight line AB is divided in the point C, so that AC:CB :: 2:3.

Problem 148.

To divide any given straight line AB in the point C, so that the whole AB is to one part AC as 5:3.

1. From point A, draw a straight line AD of indefinite

length, and at any angle to AB.

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2. Take any five equal distances on AD, and join 5 B.
3. At point 3, draw a line 3 C parallel to 5 B (Pr. 9). Then

the given straight line AB is divided in the point C, so
that AB: AC::5:3.

Section X.-SIMILAR FIGURES.

DEFINITIONS.

1. “Similar rectilineal figures are those which have their several

angles equal, each to each, and the sides about the equal angles proportionals” (Euc. VI., Def. 1).

NOTE 1.—The following rectilineal figures are similar, viz. :All equilateral triangles, squares, and regular polygons of the same

name.

NOTE 2.Other rectilineal figures, e.g., triangles which are not equilateral, trapeziums, and irregular polygons, can be made similar to given ones, as shown in the problems following.

Problem 149.

To inscribe within and equidistant from the sides of a given triangle ABC, a similar triangle, one of whose sides is equal to a given line D.

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1. Bisect the angles of the given triangle (Pr. 4) by lines

meeting in Ě.

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2. Make BF.equal to D, and through F draw FG parallel

to BE (Pr. 9).

3. Through G, draw lines parallel to the sides of the given

triangle, cutting the bisecting lines in H and K.
4. Join HK; then GHK is a similar triangle, inscribed

within the given triangle ABC, and having its side KG
equal to the given line D.

Problem 150.

To describe about and equidistant from the sides of a given triangle ABC, a similar triangle, one of whose sides is equal to a given line D.

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1. Bisect the angles of the given triangle (Pr. 4) by lines

meeting in E.
2. Produce the side BC to F, making BF equal to the given

line D.
3. Through F, draw FG parallel to the bisecting line BE,

cutting EC produced in G.
4. Through G, draw the line GH parallel to BC (Pr. 9),

cutting EB produced in H.
5. Through G and H, draw the lines GK and HK, parallel

to the sides of the given triangle ABC, meeting EA
produced in K. Then GIIK is a similar triangle
described about the given triangle ABC, and having its
side HG equal to the given line D.

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