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. D АН B G K C 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; Hlis 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 A D E F B cut the circumference in E. Join AE, EB. Draw BF parallel to AE, and join AF. FE is the rectangular parallelogram required. 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. GDB is 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. D A E H L B G F 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 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: 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 A B I D E F at the points D and F make the angles FDE, DFE equal to two of the angles of the given triangle to which the inscribed one is to be similar, .. the angle at E will be equal to the third angle. Join AE, and produce it to G; and from G draw GH, GI respectively parallel to ED, EF; join HI. HIG is the triangle required. Since DE and EF are respectively parallel to HG, GI, the angle DEF is equal to HGI. Also DE: HG :: AE: AG: EF: GI whence (Eucl. vi. 6.) the triangles HG1, DEF are similar, and.. HGI is similar to the given triangle. (29.) In a given equilateral and equiangular pentagon, to inscribe a square. perpen Let ABCDE be the given pentagon. Join EB; and from E draw EF dicular and equal to EB. Join AF; and from G, where it cuts ED, draw GH parallel to FE. Draw HI, GK parallel to EB. Join IK. HK is the square required. Since HG is parallel to EF and HI to EB, K but EF=EB, .. HG=HI. And since AE=AB, .. (Eucl. vi. 2.) HE=IB; also GK and DC being parallel to EB, and DE = BC, .. EG= BK. The triangles EHG, IKB, therefore have two sides in each and the included angles equal, and.. HG=IK, and the angle EHG=BIK, whence HG and IK are also parallel; therefore also GK is equal to HI; hence the four sides are equal; and the angle at Hbeing a right angle, all the angles are right angles, and consequently HK is a square. (30.) In a given triangle to inscribe a rhombus, one of whose angles shall be in a given point in the side of the triangle. Let ABC be the given triangle, and D the given point. Join BD, and produce it; and with the centre A, and radius AC, describe a circle cutting it in E. Join AE; and draw DF parallel to it, FG parallel to AC, and GH to FD. FH is the rhombus required. F G A C D E Since FD is parallel to AE, BF: FD :: BA: AE; and since FG is parallel to AC, BF: FG: BA AC: BA: AE, .. FD=FG; and the sides opposite to these are equal, .. the figure FDHG is a rhombus. (31.) To inscribe a circle in a given quadrant. B H G Let ABC be the given quadrant. Bisect the angle ACB by the line CD; and at D draw DE touching the quadrant, and meeting CA produced in E. Make CF-AE. From F draw FG at right angles to AC. G is the centre of the circle required. F From G draw GH perpendicular to BC. Join DF. Since the angle DCE is half a right angle, and the angle at D a right angle, DE=DC=AC=FE, .. the angle EDF=EFD; whence also GDF=GFD, and GD=GF; and since the angles FCG, GCH are equal, and GC common to the right-angled triangles GFC, GHC, .. GF=GH; .. the three lines GD, GF, GH CC |