A For the angle COD=COE, and CDO D CEO, since each of them with DOE make angles equal to two right angles, and CO is common, .. CE CD. And by con = E B struction CE=OD, and OE = CD, .. 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. E 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 B Join OG. Since EG is parallel to AB, the angle GOH= EOF, and the angles at H and Fare right angles, and GO=OE, .. HO=OF. 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 = 2 FO, FE' =4 OF2, and OE2 = 5 OF2; and if FK be drawn perpendicular to OE, OE : OK: 5 1. (22.) To inscribe a square in a given segment of circle. L 0 M C 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 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 I A B EG 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 to LH, .. KH is a parallelogram. Also since GEF is half a right angle, and EGF a right angle, . EFG is half a right angle, and .. equal to GEF; whence EG= GF, and FH-IG. But KF is equal and parallel to IG (Eucl. i. 33); .. the four sides are equal; and GFK is a right angle, .. the figure is rectangular (Eucl. i. 46. Cor.), and consequently is a square. (24.) In a given segment of a circle to inscribe a rectangular parallelogram, whose sides shall have a given ratio. АН B G Let ABC be the given segment of a circle. From A draw AD perpendicular to AC, and make AD: AC in the ratio of the sides. Complete the parallelogram AE. Bisect AC in G, and join DG; and from F draw FH perpendicular, and FI parallel to AC. Draw IK parallel to FH; HIis the rectangular parallelogram required. Since FH is perpendicular to AC, it is parallel to AD; and .. FH: HG :: AD : AG, whence FH: HK :: AD : AC, i. e. in the given ratio. And FHG being a right angle all the angles of the figure are right angles. (25.) In a given circle to inscribe a rectangular parallelogram equal to a given rectilineal figure. Let AEB be the given circle; on the diameter AB describe a rectangular parallelogram ABCD equal to the given rectilineal figure; and let the side DC F D cut the circumference in E. parallel to AE, and join AF. parallelogram required. Join AE, Draw BF FE is the rectangular For the angle EBF is equal to the two angles EBA, ABF, i. e. to EBA, BAE, and .'. is a right angle; and the angles BEA and BFA are right angles, by construction; .. also EAF is a right angle; the figure AFBE is therefore rectangular; and it is double of ABE, and (Eucl. i. 41.) .. equal to ABCD, i. e. to the given rectilineal figure. The given figure must not exceed the square of half the diameter. (26.) In a given segment of a circle to inscribe an isosceles triangle, such that its vertex may be in the middle of the chord, and the base and perpendicular together equal to a given line. Let ABC be the given segment. Bisect AC in D, and draw DE at right angles to AC, and equal to the given line. Make DF the half of DE; and join EF, meeting the circle in G. Draw GB parallel to AC; join GD, DB. the triangle required. FA GDB is E B H Since GB is parallel to AC, it is bisected by DE. Also (Eucl. vi. 2.) EH is double of GH, and .. equal to GB; .. GB and HD together are equal to ED, i. e. to the given line; and since GH=HB, and the angles at H are right angles, GD=DB, .. the triangle is isosceles. If EF does not meet the segment, the problem is impossible. When the line FE cuts the segment, there are two isosceles triangles DGB, Dgb that will answer the conditions; when it touches the segment, only one. (27.) In a given triangle to inscribe a parallelogram similar to a given parallelogram. Let ABC be the given triangle. In AB take any point D, and draw DF parallel to AC; and make the angle FDE equal to one angle of the parallelogram, and take DF: DE in D B G H F A E1 K the ratio of the sides. Join AF, and produce it to G; draw GH, HI respectively parallel to FD, DE; and GK parallel to HI. HK is the parallelogram required. For HI being parallel to DE, and HG to DF, HI: DE :: HA : DA :: HG : DF, ..HI : HG :: DE : DF, i. e. in the ratio of the sides. Also the angle GHI=FDE=one of the angles of the parallelogram, .. HIK will be equal to the adjacent angle of the parallelogram, and HK is similar to the given parallelogram. (28.) In a given triangle to inscribe a triangle similar to a given triangle. Let ABC be the given triangle, in which the triangle is to be inscribed. In AB take any point D, and draw any line DF to the opposite side; and D B G E F |