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2°. Let AH be a given straight line, and let it be required to divide it into any number of equal parts, say five. From one extremity A,

DE draw the indefinite line AG; take Al equal to any convenient line, and lay off

M IK, KL, LM, and MB, each equal to AI. Draw BH, and from I, K, L, and M, draw the lines IC, KD, LE, and MF, parallel to BH: then AH is divided into equal parts at C, D, E, and F (P. XV., C. 2).

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PROBLEM II.

To construct a fourth proportional to three given straight lines.

C

Let A, B, and C, be the
given lines.
Draw DE and

Х
В.

AH
DF, making any convenient

BH angle with each other. Lay

A off DA equal to A, DB equal to B, and DC equal to C; El draw AC, and from B draw BX parallel to AC: then DX is the fourth proportional required.

For (P. XV., C.), we have,

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Cor. If DC is made equal to DB, DX is a third proportional to DA and DB, or to A and B.

PROBLEM III.

To construct a mean proportional between two given straight

lines.

Let A and B be the given lines. On an indefinite line, lay off DE equal to A, and EF equal to B; on DF as a diameter describe the semicircle DGF, and draw EG perpendicular to DF: then EG is the mean proportional required.

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BH

AH

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To divide a given straight line into two such parts, that

the greater part shall be a mean proportional between the whole line and the other part.

D

Let AB be the given line.

E At the extremity B, draw BC perpendicular to AB, and make it equal to half of AB.

With C as a centre, and CB as a radius, describe the arc DBE; draw AC, and produce it till it terminates in the concave arc at E; with A as centre and AD as radius, describe the arc DF: then AF is the greater part required.

A

B

For, AB being perpendicular to CB at B, is tangent to the arc DBE: hence (P. XXX.),

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But, DE is equal to twice CB, or to AB: hence, AE – AB is equal to AD, or

AD, or to AF; and AB - AD is equal to AB – AF, or to FB: hence, by substitution,

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Scholium. When a straight line is divided so that the greater segment is a mean proportional between the whole line and the less segment, it is said to be divided in extreme and mean ratio.

Since AB and DE are equal, the line AE is divided in extreme and mean ratio at D; for we have, from the first of the above proportions, by substitution,

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Tu er primeradi. Yn masih
Let ABCDE lettere in
Draw CA: prolue EA, and draw

6.
BG parallel to CA; drar med sine CG.
Then the triang.es BAC and GAC hare
the common base AC, and because

5 their vertices B and G lie in the same line BG parallel to the base, their altitudes aurre equal, and consequently, the triangles are equal: hence the poligon GCDE is equal to the polygon ABCDE.

Again, draw Ce; produce AE and draw DF parallel to CE; draw also CF; then will the triangles FCE and DCE be equal: hence, the triangle GCF is equal to the polygon GCDE, and consequently, to the given polygon. In like manner, a triangle may be constructed equal to any other given polygon.

PROBLEM VII.

To construct a square equal to a given triangle. Let ABC be the given triangle, AD its altitude, and BC its base.

Construct a mean proportional between AD and half of BC (Prob. III.). Let XY be that mean proportional,

Х and on it, as a side, construct a square: then this is the square required. For, from the construction,

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Scholium. By means of Problems VI. and VII., a square may be constructed equal to any given polygon.

PROBLEM VIII.

В.

On a given straight line, to construct a polygon similar to

a given polygon. Let FG be the given line, and ABCDE the given polygon. Draw AC and AD.

At F, construct the angle GFH equal to BAC, and at

Н. G the angle FGH equal to ABC; then FGH is similar to ABC (P. XVIII. C.). In

E like manner, construct the triangle FHI similar to ACD, and FIK similar to ADE; then the polygon FGHIK is similar to the polygon ABCDE (P. XXVI., C. 2).

A

D

K

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