the number of given squares, one equal to their sum may be found.

COR. Hence lines may be found, which have the same ratio as the square roots of the natural numbers.

(10.) Having given the difference between the diameter and side of a square; to describe the square.

D

H

Let AB be the given difference. Draw AC, AD, each making half a right angle with AB; and complete the square EF. Take AG=the difference between BA and BF. Since the ratio between the side of a square and its diameter is given, that of their difference to the diameter is also given. Take.. AH : AB :: AB : AG; and through H draw HC, HD perpendicular to AC, AD; CD is the square required.

For DC being a parallelogram is also (Eucl. i. 46. Cor.) rectangular; and the angle DAH being half a right angle, is equal to DHA, . DA=DH; whence the sides are equal; and the figure is a square. And since BG= BF, HB=HC; and AB is the difference between the diameter and side.

(11.) To divide a circle into any number of concentric equal annuli.

Let ABC be the given circle, and O its centre. Draw any radius OA, and divide into the given number

DEFG

of equal portions in the points D, E, F, G, &c. On OA describe a semicircle, and draw the perpendiculars DH, EI, FK, GL, &c. Join OH, 01, &c. and with the centre O and radii OH, OI, &c. let circles be described; they will divide the circle ABC as is required.

Since the areas of circles are in the duplicate ratio of their radii; the area of the circle whose radius is OA is to that whose radius is OH in the duplicate ratio of OA: OH, i. e. in the ratio of OA: OD; .. the area of the first annulus will be to the area of the circle whose radius is OD: AD: OD. And in the same manner the area of the second annulus, will be to the area of the circle whose radius is OD, as DE: OD; and since AD=DE, the annuli will be equal. The same may be proved of

all the rest.

COR. The construction will be nearly the same, if it be required to divide the circle into annuli which shall have a given ratio; by dividing the radius AO in that proportion.

(12.) In any quadrilateral figure circumscribing a circle, the opposite sides are equal to half the perimeter. Let ABCD be a quadrilateral figure circumscribing the circle EFG; its opposite sides are equal to half the perimeter. For (Eucl. iii. 36. Cor.) AE = AH, and DH DG, .. AD is equal to AE and

=

A

H

DG together. In the same way BC is equal to BE and GC together, .. AD and BC together are equal to AB and DC together.

(13.) If the opposite angles of a quadrilateral figure be equal to two right angles, a circle may be described about it.

Let ABCD be a quadrilateral figure, whose opposite angles are equal to two right angles.

Join BD; then if a circle be described about the triangle BCD it will pass through

A. For the angle BCD and the angle in the segment BED, are together equal to two right angles, and .. equal to BCD, BAD; whence BAD is equal to the angle in the segment BED; and .. A must be a point in the circumference; or the circle will be described about ABCD.

(14.) A quadrilateral figure may have a circle described about it, if the rectangles contained by the segments of the diagonals be equal.

Let ABCD be a quadrilateral figure, the rectangles contained by the segments of whose diagonals are equal, viz. the rectangle AE, EC, equal to BE, ED.

E

D

Describe a circle about the triangle ABC; if it does not pass through D, let it cut BD in F; then (Eucl. iii. 35.) the rectangle BE, EF, is equal to the rectangle AE, EC, i. e. to the rectangle BE, ED, by the supposition; whence EF is equal to ED, the less to the greater, which is impossible; .. the circle must pass through D.

(15.) If from any point within a regular figure circumscribed about a circle perpendiculars be drawn to the sides; they will together be equal to that multiple of the semidiameter, which is expressed by the number of the sides of the figure.

Let ABCD be a regular figure circumscribed about the circle; and from any point O, let perpendiculars OE, OF, OG, &c. be drawn. Take S the centre of the circle. Join SD, SC, SH. Then the figure

will be divided into as many triangles round S and O, as there are sides of the figure; now the triangle SCD : OCD :: SH: OG; and the same being true of the triangles on each side, the sum of the triangles round S, will be to the sum of the triangles round O, as the sum of the lines SH to the sum of the perpendiculars from 0. And the first term of the proportion being equal to the second, the sum of the perpendiculars from O is equal to that multiple of the radius which is expressed by the number of the sides; each perpendicular from S being a radius of the circle.

(16.) If the radius of a circle be cut in extreme and mean ratio; the greater segment will be equal to the side of an equilateral and equiangular decagon inscribed in that circle.

Let AO, the radius of the circle ABC, be cut in extreme and mean ratio in D; AD is equal to the side of an equilateral and equiangular decagon inscribed in the

A

circle. In the circle place AC equal to AD; join CO., Then (Eucl. iv. 10.) the angles at A and C are double the angle at 0; whence AOC is one fifth part of two right angles, or one tenth part of four right angles, i. e. of the angles at 0; and .. AC is the side of a regular decagon inscribed in the circle.

(17.) Any segment of a circle being described on the base of a triangle; to describe on the other sides segments similar to that on the base.

Let ABC be a triangle, on the base AC of which a segment of a circle ADC is described. Produce AB, CB to E and D. Join AD, CE; and through A, D, B, and C, E, B let circles be described; the segments ADB, BEC are similar to ADC.

B

E

For the angle ADC being in the segments ADB, ADC, those segments are similar. For the same reason the segments ADC, BEC are similar. And since the angles ADC, AEC are equal, . the segments ADB BEC are similar.

(18.) If an equilateral triangle be inscribed in a circle; the square described on a side thereof is equal to three times the square described upon the radius.

Let ABC be an equilateral triangle inscribed in a circle. From A draw the diameter AD, and take O the centre;

BB