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HAving given the perpendicular of a right-angled Triangle that is to be let fall from the right angle, and the diffe∣rence of the segments of the base, to find the segments, and de∣scribe the Triangle.
E. g. If the perpendicular is, as above, BE, and the diffe∣rence of the segments AH (n. 1. Fig. 38.) to find the seg∣ments AD and DB, from whose common term you are to e∣rect a perpendicular DG or DC to form the Triangle.
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Make the lesser segment =x, and the difference of the seg∣••ts =a, the greater segment will be x+a Make the gi∣•• perpendicular as before =b: Therefore you'l have as x+a to b so b to x; and consequently, xx+ax=bb; and substracting ax, xx=bb−ax. Wherefore according to case 2. 〈 math 〉〈 math 〉.
The Geometrical Construction. Make HD=½a, DG equal perpendicular to b; HG will be = 〈 math 〉〈 math 〉 = HB or ••, viz having drawn a semi-circle from H thro' G. ••erefore DB is the less segment, and AD the greater; and ••ing drawn AG and BG, or on the other side, (making the ••pendicular DC=DG) having drawn AC and BC, the ••iangle will be constructed. Or with Cartes, make (n. 2.) ••=½a and EB=b, and having described a Circle from ••hro' E draw BHA; and so you'l have the two segments ••ght AB the greater and DB the lesser.
The Arithmetical Rule. Join the squares of the half diffe∣••ce and perpendicular into one sum, and then having extra∣•• the root substract half the difference from it; and the re∣••nder will be the lesser segment sought; and having added •• difference you'l have also the greater.