one of the given lines. Again, since GB is parallel to AC, (Eucl. vi. 2.) it bisects AE, and BG='AC; but DB= == AC, .. DG=AC, and .. is equal to another of the given bisecting lines. In the same manner, AF may be shewn to be equal to the other given line, and to bisect DE in F. (36.) Given the segments of the base made by the perpendicular, and one of the angles at the base triple the other ; to construct the triangle: ' Let AB, BC equal to the given segments, be placed in the same straight line. Make BD=BC; bisect AD in E, and BĘ in F. On AD describe a semicircle ; and from Fdraw FG at right angles to AD. Join AG, GD; and let AG meet the perpendicular BH in H. Join HC; AHC is the triangle required. Draw EI perpendicular to AD; join DI, DH. Then AE being equal to ED and the angles at E right angles, Al= ID, and the angle IAD= IDA; whence the angle DIH is double of DAH. But since EF is equal to FB, and GF parallel to IE and BH, .. IG= GH; and the angles DGI, DGH being right angles, DI is equal to DH,' and the angle DHI equal to DIH, and .. double of DAH. Also since DB=BC, and the angles at B are right angles, :. the angle HCB is equal to HDB, i. e. to DHA, DAH together, or to three times the angle DAH. Also AB, BC, by construction, are equal to the given segments made by the perpendi. cular. (37.) The area and hypothenuse of a right-angled triangle being given ; to describe the triangle. Let AB be equal to the given hypothenuse. Bisect it in C, and on CB describe a rectangular parallelogram CD equal to the given area. On AB describe a semicircle AEB, cutting the side parallel to AB, in E. Join AE, EB; AEB is the triangle required. Join CE. Since AC = CB, the triangle AEB is double of CEB (Eucl. i. 38.), and .:. equal to the rectangle CD, i.e. to the given area. B a line; and upon (38.) Given one angle, and a line drawn from one of the others bisecting the side opposite to it ; to construct the triangle, when the area is also given. Let AB be the given bisecting it describe a segment of a circle containing an ångle equal to the given angle. Upon AB also describe a rectangular parallelogram ABCD equal to the given area ; and let DC meet the circle in E ; join EA, and produce it, making AF=AE; join FB, BE; FEB is the triangle required. For BA bisects the side EF; the angle BEF is equal to the given angle ; and the triangle BEF is double of BAE (Eucl. i. 38.) and .:. equal to ABCD, i. e. to the given area. (39.) In two similar right-angled triangles, the sum of the base of the one and perpendicular of the other is given ; to determine the triangles such that their hypothenuses may contain the right angle of another triangle similar to them, and the sum of the three areas may be equal to a given area. D B Let AB be equal to the given sum; and upon it describe a rectangular parallelogram equal to the given area. On AD also describe a semicircle, cutting CD in E; join AE, EB; the triangles AED, BEC, AEB are the triangles required; as is evident from the construction. a (40.) Given the vertical angle, the area, and the distance between the centres of the inscribed circle and the circle which touches the base, and the two sides produced; to construct the triangle. h Let AB be equal to the given distance between the centres; and make the angle BAC equal to half the given angle at the vertex, On AC let fall the perpen dicular BC; and produce CA, till the rectangle AD, DC is to the given area, in the ratio of AC : CB. Complete the parallelogram DEFC; and from the centrés A and B, describe two circles touching EF in G and F. Draw HI, IE touching the two circles; HIE is the triangle required. For AB is equal to the given distance between the centres. The angle IEF is double of AEF, i. e. of BAC, and .. is equal to the given angle. And from the similar triangles ABC, AED, AC : CB :: AD : (DE=) AG :: ADx DC : AG [~ EF. But since EF is equal to half the perimeter of the triangle EHI, the rectangle AG, EF is equal to the triangle EHI; whence EHI is equal to the given area. (41.) Given the area, the line from the vertex dividing the base into segments which have u given ratio, and either of the angles at the base ; to construct the triangle. G K B On AB the given line, describe a segment of a circle containing an angle equal to the given angle. Describe a rectangular parallelogram ABCD equal to twice the given area; and divide BC in the given ratio at F. Through F draw GH parallel to AB. Join BH, HA; and produce HB so that IB may be to BH in the given ratio. Join IA; IAH will be the triangle required. Draw IK perpendicular to AB. The triangles IKB, BFH are similar, . . IK : BF :: IB : BH :: FC : FB, ::. IK=FC; and the triangle 1 AB is equal to half of the parallelogram GC; and the triangle ABH is equal to half of BG; ;. IAH is equal to half the parallelogram AC, and .. equal to the given area. And AB is equal to the given dividing line, and IB : BH in the given ratio. B G FA E (42.) Given the difference between the segments of the base made by the perpendicular, the sum of the squares of the sides, and the area ; to construct the triangle. Take a line AB such that its square may be equal to the difference between half the given sum of the squares, and the square of half the given difference of the segments of the base. Upon AB describe a rectangular parallelogram ABCD equal to the given area ; and also on AB describe a semicircle cutting CD in E. Join AE, and produce it both ways; and make AF, AG, each equal to half the given difference of the segments, and make EH=EG. Join BF, BH; BFH is the triangle required. Join BG, BE. Since GE=EH and the angles at E are right angles, :. GB=BH; and the sum of the squares of FB, BH is equal to the sum of the squares of FB, BG, i.e. (iv. 30.) is equal to twice the sum of the squares of FA and AB, i.e. by construction, is equal to the given sum Also the difference between FE and EH is equal to the difference between FE and EG, i. e. to FG, which is equal to the given difference. And the area of the triangle FBH is double of the area of the triangle ABE, and :: (Eucl. i. 41.) equal to ABCD, or the given area. |