Elements of geometry and trigonometry . to the sunt of the scjuares uponJ:D and EG. Secondli/. If it is required to find a square equivalent to thedifference of the given s(iuares,fonniii the same manner the rightangle FEH ; take GE equal to tlie shorti-r of the sides A andB ; fn/in th«i point G as a centre, willi a radius Gif, equal tothe other side, describe an arc cutting EH in 11 : tlie stjuaredescribed upon Ell will be equivalent to the différence \ji the«quares described upon llie lines A and B. JTor the triangle GEII is right aiigk-d, the hypothonuscGil = A, and the side GE —B; hence ih
Elements of geometry and trigonometry . to the sunt of the scjuares uponJ:D and EG. Secondli/. If it is required to find a square equivalent to thedifference of the given s(iuares,fonniii the same manner the rightangle FEH ; take GE equal to tlie shorti-r of the sides A andB ; fn/in th«i point G as a centre, willi a radius Gif, equal tothe other side, describe an arc cutting EH in 11 : tlie stjuaredescribed upon Ell will be equivalent to the différence \ji the«quares described upon llie lines A and B. JTor the triangle GEII is right aiigk-d, the hypothonuscGil = A, and the side GE —B; hence ihe s(jiiare describedU|Kjn EH, is eciuivalent to the difference of the squares AaiKlB. Scholium. A s(piare may thus be found, e(|uivalent to thesum of any number of stpiares ; for a similar construction whichreduce-s two of them to one, will reduci- three «Tf them to two,and these two to one, and so of others. It would be tin* sam»\if any of the squares were to be subtracUd from the sum of•the others. 104 GEOMETRY. PROBLEM To find a square which shall he to a given square as a givenline to a given line. Let AC be the givensquare, and M and N thegiven hnes. Upon the indefiniteline EG, take EF=M,and FG=N ; upon EGas a diameter describea semicircle, and at the point F erect the perpendicular the point II, draw^ the chords HG, HE, which produceindefinitely : upon the first, take HK equal to the side AB ofthe given square, and through the point K draw KI parallel toEG ; HI will be the side of the square required. For, by reason of the parallels KI, GE. we have HI : HK: : HE : HG; hence, HP : HK^ : : HE^ : HG^: but in theright angled triangle EHG, the square of HE is to the squareof HG as the segment EF is to the segment FG (Prop. 3.), or as M is to N; hence HP : HK^ . m : N. ButIIK=AB ; therefore the square described upon HI is to thesquare described upon AB as M is to N. PROBLEM XIL Upon a given line, to describe a polygon similar to a given polygon. Le
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