=

Dem. - Since BE is equal to GF, the rectangle BE. GE BF. GF. Therefore DE. CE DF. AF; hence DE: DF

:: AF: CE; and by similar triangles, AB: AF:: DE: DF, and

CE:CB:: DE: DF. Hence AB: AF :: AF: CE, and AF: CE :: CE CB. Therefore AB, AF, CE, CB are continual proportions. Hence [vi. Def. Iv.] AF, CE are two mean proportionals between AB and BC.

The foregoing elegant construction is due to the ancient Geometer PHILO of BYZANTIUM. If we join DG it will be perpendicular to EF. The line EF is called Philo's Line; it possesses the remarkable property of being the minimum line through the point B between the fixed lines DE, df.

NEWTON'S CONSTRUCTION.-Let AB and L be the two given

lines of which AB is the greater. Bisect AB in C. With A as centre and AC as radius, describe a circle, and in it place the

chord CD equal to the second line L. Join BD, and draw by trial through A a line meeting BD, CD produced in the points E, F, so that the intercept EF will be equal to the radius of the circle. DE and FA are the mean proportionals required.

Dem.-Join AD. Since the line BF cuts the sides of the A ACE, we have

AB. CD. EF CB.DE. FA; but EF = CD;

Again, since the AACD is isosceles, we have

ED. EC EA? - AC2 = (FA + AC)2 - AC2

=

therefore DE2 FA. AB, and we have AB. CD DE. FA.

Hence

=

=

AB, DE, FA, CD are in continued proportion.

NOTE E.

ON PHILO'S LINE.

I am indebted to Professor Galbraith for the following proof of the minimum property of Philo's Line. It is due to the late Professor Mac Cullagh :-Let AC, CB be two given lines, E a fixed point, CD a perpendicular on AB; it is required to prove, if AE is equal to DB, that AB is a minimum.

Dem.-Through E draw EM parallel to BC; make EN = EM; produce AB until EP = AB. NT, RP each parallel to AC,

Through the points N, P draw and through P draw PQ parallel

to BC. It is easy to see from the figure that the parallelogram QR is equal to the parallelogram MF, and is therefore given. Through P draw ST perpendicular to EP. Now, since AE= DB, BP is equal to DB; therefore PS CD. Again, since OP = AD, PT is equal to CD; therefore PS PT. Hence QR is the maximum parallelogram in the triangle SVT.

=

=

=

Again, if any other line A'B' be drawn through E, and produced to P', so that EP' = AP', the point P' must fall outside ST, because the parallelogram QʻR', corresponding to QR, will be equal to MF, and therefore equal to QR. Hence the line EP' is greater than EP, or A'B' is greater than AB. Hence AB is a

minimum.

NOTE F.

ON THE TRISECTION OF AN ANGLE.

The following mechanical method of trisecting an angle occurred to me several years ago. Apart from the interest belonging to the Problem, it is valuable to the student as a geometrical exercise: ·

To trisect a given angle ACB.

Sol.-Erect CD perpendicular to CA; bisect the angle BCD by CG, and make the angle ECI equal half a right angle; it is evident that CI will fall between CB and CA. Then, if we use a

jointed ruler; that is, two equal rulers connected by a pivot, and make CB equal to the length of one of these rulers, and with C as a centre and CB as radius, describe the circle BAM, cutting CI in I; at I draw the tangent IG, cutting CG in G.

Then, since ICG is half a right angle, and CIG is right, IGC is half a right angle, therefore IC is equal to IG; but IC equal CB; therefore IG CB, equal length of one of the two equal rulers. Hence, if the rulers be opened out at right angles, and placed so that the pivot will be at I, and one extremity at C, the other

=

extremity at G; it is evident that the point B will be between the two rulers; then, while the extremity at C remains fixed, let the other be made to traverse the line GF, until the edge of the second ruler passes through B: it is plain that the pivot moves along the circumference of the circle. Let CH, HF be the positions of the rulers when this happens; draw the line CH; the angle ACH is one-third of ACB.

=

Dem.-Produce BC to M. Join HM. Erect BO at right angles to BM. Then, because CH=HF, the angle HCF=HFC, and the angle DCE ECB (const.). Hence the angle HCD · HBC [1. XXXII.], and the right angles ACD, CBO are equal; therefore the angle ACH is equal to HBO; that is [III. XXXII.], equal to HMB, or to half the angle HCB. third of ACB.

=

Hence ACH is one

NOTE G.

ON THE QUADRATURE OF THE CIRCLE.

Modern mathematicians 'denote the ratio of the circumference of a circle to its diameter by the symbol π. Hence, if r denote the radius, the circumference will be 2πr; and, since the area of a circle [vi. xx. Ex. 15] is equal to half the rectangle contained by the circumference and the radius, the area will be r2. Hence, if the area be known, the value of π will be known; and, conversely, if the value of π be known, the area is known. On this account the determination of the value of π is called "the problem of the quadrature of the circle," and is one of the most celebrated in Mathematics. It is now known that the value of π is incommensurable; that is, that it cannot be expressed as the ratio of any two whole numbers, and therefore that it can be found only approximately; but the approximation can be carried as far as we please, just as in extracting the square root we may proceed to as many decimal places as may be required. The simplest approximate value of π was found by Archimedes, namely, 22:7. This value is tolerably exact, and is the one used in ordinary calculations, except where great accuracy is required. The next to this in ascending order, viz. 355: 113, found by Vieta, is correct to six places of decimals. It differs very little from the ratio 3-1416 1, given in our elementary books.

:

Several expeditious methods, depending on the higher mathematics, are known for calculating the value of . The following is an outline of a very simple elementary method for determining this important constant. It depends on a theorem which is at once inferred from vi., Ex. 87, namely, "If a, A denote the reciprocals of the areas of any two polygons of the same number of sides inscribed and circumscribed to a circle, a', A' the corresponding quantities for polygons of twice the number, a' is the geometric mean between a and A, and A' the arithmetic mean between a' and A.” Hence, if a and A be known, we can, by the processes of finding