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PROBLEMS OF CONSTRUCTION.

G

B

Proposition 33. Problem. 345. To divide a given straight line into parts proportional to given straight lines.

Given, the line AB, and the lines P NM
M, N, P.
Required, to divide AB into parts

F proportional to M, N, P. Cons. From A draw the indefinite

G straight line AH making any _ with

D
AB.
On AH take

E!
AC = M, CD=N, DE = P.

IH
Join EB, and through D, C draw
DG, CF || to EB.

Then AB is divided at G and F into parts proportional to M, N, P. Proof. Since CF, DG, EB are II,

(Cons.) AF FG GB

AC CD DE If two st. lines are cut by any number of || 8, the corresponding segments

are proportional (300). But AC = M, CD = N, DE = P.

(Cons.)
AF FG GB
M = N =
P.

Q.E.F. 346. Sch. If it be required to divide a line AB into any number of equal parts, take the same number of equal parts on AH, so that AC = CD = DE, and complete the construction as before; then AF = FG = GB.

EXERCISES. 1. Divide a straight line into five equal parts.

2. Give the construction for cutting off two-sevenths of a given straight line.

Proposition 34. Problem. 347. To find a fourth proportional to three given straight lines.

D

Given, the three lines M, N, P.

M

N Required, to find a fourth propor

P tional to M, N, P.

Cons. Draw the two indefinite straight lines AG, AH, making any _ with each other. On AH take AB =

B : M, BC= N; and on AG take AD = P.

C Join DB, and through C draw CE || to

H н DB. Then DE is the fourth proportional

E required. Proof. Since BD and CE are II,

.. AB: BC = AD: DE. A st. 1. || to a side of a A divides the other two sides proportionally (298). But

AB = M, BC = N, AD = P. (Cons.) .::M:N=P:DE.

Q. E.F. 348. Sch. If the lines N and P are equal, then BC and AD are both laid off equal to N, and DE is the third proportional to M and N (280). The proportion in (347) then becomes

M:N=N:DE.

EXERCISES.

1. Construct a fourth proportional to the lines 3, 7, 11.

2. If from D, one of the angles of a parallelogram ABCD, a straight line is drawn meeting AB at E and CB produced at F; show that CF is a fourth proportional to EA, AD,

and AB.

Proposition 35. Problem.

N

349. To find a mean proportional between two given straight lines. Given, the lines M and N.

M Required, to find a mean proportional between M and N.

Cons. On an indefinite straight line take AD = M, and DB = N. On AB describe a semicircle, and

A draw DC I to AB.

Then DC is the mean proportional required.

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350. Sch. The mean proportional between two lines is often called the geometric mean, while their half sum is called the arithmetic mean,

EXERCISE,

From a given point P in a circle a perpendicular PM is drawn to a given chord AB; from A, B perpendiculars AC, BD are drawn to the tangent at P: prove that PM is a mean proportional between AC and BD.

351. DEF. A straight line is said to be divided in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less.

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

A
AB : AC = AC: CB.

D

F

= AB

Proposition 36. Problem. 352. To divide a given straight line in extreme and mean ratio.

Given, the st. line AB.

Required, to divide it in extreme and mean ratio.

Cons. At B erect the BC==} AB.
With centre C and radius CB de-

A scribe a o.

Join AC, cutting the o at D, and produce it to meet the O again at E. On AB lay off

AF = AD. Then F is the pt. of division required. Proof. Since AB is a tangent and AE a secant to the o, .:. AE: AB = AB: AD.

(1) The tangent is a mean proportional between the whole secant and the

external segment (339). .. by division, AE · AB: AB

AD: AD. (288) But AB= DE and AD= AF.

(Cons.) .:. AE – AB = AD AF, and AB AD= FB. Substituting these values in the above proportion, we have AF: AB = FB: AF.

(285) AB: AF = AF: FB. .:. AB is divided in extreme and mean ratio at F. Q. E.F.

353. Sch. If BA be produced to the left of A to a point F' so that AF' = AE, then F' is another point of division having the same property as F. Thus we have from proportion (1) in (352), by composition,

AE + AB : AE = AB + AD: AB. (287) But AE + AB = BF', and AB + AD = AF'. (Cons.)

... BF: AF = AF: AB, or

AB: AF AF : F'B. .. AB is divided in extreme and mean ratio at F'. AB is said to be divided at F internally, and at F' externally, in extreme and mean ratio.

NOTE.-This division of the straight line was called by the ancient geometers the golden section.

Proposition 37. Problem. 354. On a given straight line to construct a polygon similar to a given polygon.

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Cons. Let AB be the side of the given polygon to which A'B' is to be homologous; join AC and AD.

A At A' make _B’A'C' = ZBAC, and at B' make _ A'B'C' = _ ABC.

B

Also at A' make Z C'A'D' = _CAD,

and

at C' make ZA'C'D' = ZACD.

Also at A' make _D'A'E' = _DAE, and at D' make Z A'D'E' = _ADE. Then A'B'C'D'E' is the required polygon.

Proof. Since the As ABC, A'B'C' are mutually equiangular,

(Cons.) i', they are similar.

(309) Also, AS ACD, A'C'D' are similar.

(309) Likewise, As ADE, A'D'E' are similar,

(309) .. the polygons ABCDE, A'B'C'D'E' are similar, being composed of the same number of 18, similar each to each, and similarly placed (320).

Q.E.F.

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