PROP. XXXV. PROBLEM.

294. To divide a given straight line into parts proportional to any number of given lines.

Given line AB, and lines m, n, and p.

Required to divide AB into parts proportional to m, n, and p.

Construction. On the indefinite line AC, take AD=m; on DC take DE = n; on EC take EF=p; and draw line

BF.

Draw lines DG and EH || to BF, meeting AB at G and ́ H, respectively.

Then, AB is divided into parts AG, GH, and HB proportional to m, n, and p, respectively.

Proof. Since DG is || to side EH of ▲ AEH,

And since EH is to side BF of ▲ ABF,

(?)

(1)

(2)

(?)

Ex. 76. Construct a line equal to √2; to √5; to vē.

PROP. XXXVI. PROBLEM.

295. Upon a given side, homologous to a given side of a given polygon, to construct a polygon similar to the given polygon.

Given polygon ABCDE, and line A'B'.

Required to construct upon side A'B', homologous to AB,

a polygon similar to ABCDE.

Construction. Divide polygon ABCDE into A by drawing diagonals EB and EC.

At A' construct B'A'E'

=

ZA; and draw line B'E',

making A'B'E' = LABE, meeting A'E' at E'.

Then, ▲ A'B'E' will be similar to ▲ ABE.

(?)

In like manner, construct ▲ B'C'E' similar to ▲ BCE, and ▲ C'D'E' similar to ▲ CDE.

Then, polygon A'B'C'D'E' will be similar to polygon ABCDE.

($ 266)

296. Def. A straight line is said to be divided by a given point in extreme and mean ratio when one of the segments (§ 250) is a mean proportional between the whole line and the other segment.

Thus, line AB is divided internally in extreme and mean ratio at C if

AB: AC AC: BC;

and externally in extreme and mean ratio at D if

PROP. XXXVII. PROBLEM.

297. To divide a given straight line in extreme and mean ratio (§ 296).

Required to divide AB in extreme and mean ratio. Construction. Draw line BE AB, and equal to 1⁄2 AB. With E as a centre and EB as a radius, describe O BFG. Draw line AE cutting the circumference at F and G. On AB take AC = AF; on BA produced, take AD = AG. Then, AB is divided at C internally, and at D externally, in extreme and mean ratio.

Proof. Since AG is a secant, and AB a tangent,

Therefore, AB is divided at C internally in extreme and

mean ratio.

Again, from (1),

AG + AB: AG = AB+ AC: AB.

But,

AG + AB= AD + AB = BD.

And by (2), AB+ AC = FG + AF = AG.

(?)

.. BD: AG = AG: AB.

.. AB: AG =

AG: BD.

.. AB: AD = AD: BD.

(?)

Therefore, AB is divided at D externally in extreme and mean ratio.

298. Cor. If AB be denoted by m, and AC by x, proportion (3) of § 297 becomes

Multiplying by 4, and adding m2 to both members, 4x2 + 4 mx + m2 = 4 m2 + m2 = 5m2.

Extracting the square root of both members,

Since x cannot be negative, we take the positive sign before the radical sign; then,

77. To inscribe in a given circle a triangle similar to a given triangle. (§ 261.)

(Circumscribe a about the given A, and draw radii to the vertices.)

78. To circumscribe about a given circle a triangle similar to a given triangle. (§ 262.)

BOOK IV.

AREAS OF POLYGONS

PROP. I. THEOREM.

299. Two rectangles having equal altitudes are to each other as their bases.

Note. The words "rectangle," "parallelogram," "triangle,” etc., in the propositions of Book IV., mean the amount of surface in the rectangle, parallelogram, triangle, etc.

Case I. When the bases are commensurable.

Given rectangles ABCD and EFGH, with equal altitudes AB and EF, and commensurable bases AD and EH.

Proof. Let AK be a common measure of AD and EH, and let it be contained 5 times in AD, and 3 times in EH.

Drawing Is to AD and EH through the several points of division, rect. ABCD will be divided into 5 parts, and rect. EFGH into 3 parts, all of which parts are equal.

(§ 114)

(2)

(?)