(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.

D

E

B

H G

Let ABCD be a regular figure circumscribed about the circle; and from any point O, let perpendiculars OE, OF, OG, &c. be drawn. Take Sthe 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

DO

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.

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

join BD, BO. Then the angle BOD= BAC = BCA=BDO, .. BD=BO; and the squares of AB, BD are equal to the square of AD, i. e. to four times the square of BO, or BD; and ... the square of AB is

B

equal to three times the square of BD or BO.

(19.) To inscribe a square in a given right-angled isosceles triangle.

F

B

G

Let ABC be a right-angled isosceles triangle, having the side BA = BC. Trisect the hypothenuse AC in the points D, E; and from D, E draw DF, EG perpendicular to AC; join FG; DFGE is the square required.

Since the angle DAF is half a right angle, and the angle at D a right angle, .. the angle DFA is half a right angle, and equal to DAF; whence DF-DA. In the same manner it may be shewn that EG=EC. But AD=EC; and .. FD, DE and EG are equal; and (Eucl. i. 33.) FG = DE; DE; .. the figure is equilateral. And it is rectangular, (Eucl. i. 46.) since the angles at D and E are right angles; .. it is a square.

(20.) To inscribe a square in a given quadrant of

a circle.

Let AOB be the given quadrant, whose centre is O. Bisect the angle AOB by the line OC. Draw CE, CD parallel to OA, OB. DE is a square.

A

D

E B

For the angle COD=COE, and CDO = CEO, since each of them with DOE make angles equal to two right angles, and CO is common, .'. CE=CD. And by construction CE=OD, and OECD, .. the figure is equilateral. And the angle DOE is a right angle, .. (Eucl. i. 46. Cor.) all its angles are right angles; and consequently the figure is a square.

(21.) To inscribe a square in a given semicircle.

Let ACB be the given semicircle;

take its centre, and from B draw BD perpendicular and equal to BA. Join OD, cutting the circumference in E; and from E draw EF perpendicular to AB, and EG parallel to it; draw GH parallel to EF. Then EH is the square required.

H

Join OG. Since EG is parallel to AB, the angle GOH= EOF, and the angles at H and Fare right angles, and GO= OE, .. HOOF.

Now EF FO :: DB : BO,

:

.. EF=2 FO=FH; the figure is.. equilateral; and it is, by construction, rectangular, .. it is a square.

COR. Since FE=2FO, FE' = 4 OF2, and OE2 = 5 OF; and if FK be drawn perpendicular to OE, OE: OK :: 5 : 1.

(22.) To inscribe a square in a given segment of a circle.

L

A H D EF B

Let AIB be the segment of a circle, whose base AB is bisected in D. From B draw BC perpendicular to BA and equal to BD. Bisect BD in E, and join CE. Draw DG parallel to CE, and GF to CB. Take DH= DF; draw HI perpendicular to AH, and .. parallel to GF; Join GI. FI is the FI is the square required.

Since GD and GF are respectively parallel to CE and CB,

GF: FD:: CB : BE :: 2 : 1,

.. GF=2 FD= FH. Take O the centre, and draw OL, OM perpendiculars to HI, FG; then since HD=DF, OL=OM, .. (Eucl. iii. 14.), IL=GM; but LH = FM, .. IH=GF; whence IG=HF, and the figure is equilateral; and since the angle at F is a right angle, the figure is rectangular, and.. is a square.

(23.) Having given the distance of the centres of two equal circles which cut each other; to inscribe a square in the space included between the two circumferences.

K

F

A

BG

L

B H

Let A and B be the centres of two equal circles, which cut each other in C and D. Join AB, and bisect it in E; and at the point E make the angle GEF half a right angle; and from F draw FGH perpendicular to AB. Make EI= EG; and through I draw KL perpendicular to AB; join KF, LH. KH is a square.

Since EI-EG, BI= AG, and .. (Eucl. iii. 14.) KL = FH; and they are parallel, .. KF is equal and parallel