draw a second line parallel to NO meeting the given lines as at A and C. If we divide AC, as in Q, in similar segments to those in which M divides NO (Problem 3) the line QM will be the required line passing through the intersection of AB and CD. The most convenient method for dividing AC is probably thus : join CN, draw PM through M parallel to CD and meeting CN in P, and through P draw PQ parallel to AB meeting AC in Q. AC is obviously divided in similarly to NC in P and therefore to NO in M.

PROBLEM 5. (Fig. 5.) To find the geometric mean between two given lines AB, CD, i.e. to find a line of length (1) such that

AB : 1 :: 7 : CD,

or that the square on I shall be equal in area to the rectangle contained by AB and CD.

Draw any straight line EOF and set off on it on opposite sides from 0, OE = AB, OF=CD. On EF describe a semicircle

and from 0 draw OG perpendicular to EF meeting the circumference in G. OG will be the required mean proportional or geometric mean. For, since the angle in a semicircle is a rightangle (Euclid III. 31), .. the angles OEG, EGO are together equal to the angles EGO, OGF, and .. the angle OEG = the angle OGF, .. the right-angled triangles OEG, OGF are similar and

i. e.

.. EO: OG :: OG : OF,

AB : OG :: OG : CD.

PROBLEM 6. To divide a given line so that the rectangle contained by its segments is equal to the square on a given line which

must obviously be not greater than half the line to be divided (fig. 5).

This is the converse of the last problem. Let EF be the given line, on it describe a semicircle. Draw the radius KL perpendicular to EF and on it make KM equal to the side of the required square. Through M draw a parallel to EF meeting the circle in G and from G drop a perpendicular on EF meeting it in 0, 0 will be the required point of division.

The construction is obvious from the last problem.

PROBLEM 7. (Fig. 6.) To divide a line medially, or in extreme and mean proportion, i.e. to find a point (F) in a line AB such

that

the whole line AB the greater segment (BF)

:: BF: the lesser segment (AF),

or that the rectangle contained by the whole line and the lesser segment is equal in area to the square on the greater segment.

Bisect AB in C, from A draw AD perpendicular to AB and make_AD=AC = AB. Join BD and on it from D cut off

DE = DA ; from B on BA cut off BF = BE. F will be the required point. This construction is simplified from Euclid 11. 11, the proof may be shewn thus.

sq. on AB + sq. on AD (Euclid 1. 47).

The sq. on BD

ED AD and EB=FB,

sq. on EB+ sq. on ED + 2 rect. EB.ED,

sq. on AB=sq. on FB + 2 rect. FB.AD.

(Euclid 11. 4),

Again sq. on AB = sq. on FB + sq. on AF +2 rect. AF. FB

(Euclid 11. 4),

sq. on FB +rect. AF (AF+ FB)
+rect. AF. FB.

but

2 rect. FB.AD = rect. AF. AB + rect. AF.FB,

i. e. rect. FB {2AD-AF}=rect. AF. AB,
2AD = AB and AB – AF = FB,

.. finally sq. on FB= rect. AF. AB.

PROBLEM 8. (Fig. 7.) To find graphically a series of terms in geometrical progression, being given either two successive terms or one term and the common ratio.

Draw two lines Oe, OF meeting in O at any convenient angle. On one mark off the 1st given term as OA, and on the other the

2nd given term as OB, or if the common ratio be given a length OB= 1st term multiplied by the common ratio.

[In the figure OA the first term = 2, and OB = 2·4; the common ratio therefore is 1·2, the unit being 3′′/8.]

With centre O and radius OB describe an arc cutting OA in b; through b draw bC parallel to AB cutting OB in C. OC will be the required third term of the series. Similarly make Oc on OA

OC and through c draw cD parallel to AB cutting OB in D, OD will be the required fourth term, and so on in succession. Terms on the other side of OA can also be determined as shown at OB1, OC1, &c.

The construction evidently depends on the similarity of the triangles OAB, ObC, &c.

by which

i.e. since

ОС : ОВ :: ОЪ : OA,

Ob = OB, OB2 = OA. OC,

10

PRODUCT OF TWO RATIOS.

or each term is a mean proportional between the two on opposite sides of it, in other words the series is in geometrical progression. Since OB =r. OA, the above expression for OB2 becomes r2. OA = OC,

and so also

p3. OA = OD and so on.

Very careful drawing is required to ensure accuracy, and the scale should be as large as possible, as otherwise, since errors are cumulative, the lengths obtained for the fourth or fifth and succeeding terms may differ considerably from their true values.

al

bm

PROBLEM 9. Given two ratios

a

Given two ratios & and to determine the ratio

m

or to divide a given line so that the ratio of its segments shall equal the product of two given ratios (Fig. 8).

Draw any line AB and on it make AD=a, DB=b. With centre B and radius l+m describe an arc, and with centre

A and radius AC the length of the given line to be divided describe an arc intersecting the former in C. Make BF on BC = 1 so that FC = m. Draw AF, CD intersecting in O and draw BO E will be the required point of division :

meeting AC in E.

It follows of course that in any triangle if lines be drawn from the vertices ABC meeting the opposite sides in F, E, D and all passing through the same point O, AD. BF. CE = DB . FC . EA, i.e. that the continued products of the alternate segments taken in order are equal.

PROBLEM 10. To determine graphically the square root of any number (n), i.e. to determine a line the length of which length of a line containing n units measured on any scale :: 1 : √n. This is sometimes, though misleadingly, called determining the square root of a given line. The fact is that the expression the square root of a given line has no meaning unless we take the line to represent, by the number of units it contains, a given area; and then the line to be found is the side of a square, the number of square units in which is equal to the number of units contained in the line—the same scale of course being used for each. If a triangle ABC be drawn, right angled at A and having the sides AB, AC each one inch long, the side BC is the side of a square of two square inches area, and in this sense BC may be said to be the square root of a line two inches long, or of the number 2, the unit being one inch, but if the unit be half-an-inch the same line BC represents the square root of 8, since (Euc. 1. 46) BC2 = AB2 + AC2 = 22 + 22 = 8.

If we use a diagonal scale of half-inches, the length BC may be read on it to two places of decimals, and the number so obtained is the square root of 8 to two decimal places. Any question relating to the square root of a number, must therefore always be taken as involving the application of some particular scale. The square root of any proposed number can be found by splitting the number up so as to make it equal to the sum or difference of two or more squares, and then constructing right-angled triangles having sides equal to the sides of these