But the Trapezium A B x C + A C x D = □ ABCD and the Trapezium ab x D + ▲ Cx Dab CD. consequently, A B C D □ a b CD. Q. E. D. Corollary. Hence it will be easy to conceive, that all Triangles which stand upon the fame Bafe, or upon equal Bafes, and between the fame Parallels, (viz. having the fame Height) are equal one to another. (37 & 38. e. 1.) For all Triangles are the Halfs of their circumfcribing Parallelograms; and therefore, if the Wholes be equal, their Halfs will al fo be equal. THEOREM XXI. Parallelograms (and confequently Triangles) which have the fame Height, have the fame Proportion one to another as their Bafes have. (1. e.. 6.) Like Triangles are in a duplicate Ratio to that of their homologous Sides. (19. e. 6.) That is, the Area's of like Triangles are in Proportion one to another as are the Squares of their like Sides. Demonftration. Suppofe the ABCD and Abcd to be alike, and their like Sides to be those mark'd with the fame Letters. B Ꭰ Let Let A and a be Perpendiculars to the two Bafes D and d. Then DA= the Area of A B C D } By Lemma 3, Page 303. And d a = the Area of A b c d 2 But IB:b:: D: d / DD da DddA. By Axiom 3. 2 THEOREM XXIII. Q. E. D. In every Obtufe-angled Triangle (as B CD) the Square of the Side fubtending the obtufe Angle (as D) is greater than the Squares of the other two Sides (B and C) by a double Rectangle made out of one of the Sides (as B) and the Segment or Part of that Side produced, (as a) until it meet with the Perpendicular (P) let fall upon it. (12. e. 2.) That is, DDBB+C C+ 2 Ba. Hence 'tis evident, that, if the Sides of any Obtufe-angled Triangle are given, the Segment (a) of the Side produced (or the Perpendicular, P) may be easily found. THEOREM XXIV. If a Perpendicular (as P) be let fall in any Acute angled Triangle (as BCD), the Square of either of the Two Sides (as D) is lefs than the Squares of the other Side, and that Side upon which the Perpendicular falls (viz. C and B) by a double Rectangle made of the Side B, and that Segment or Part of it (viz. a) which lies next to the Side C. (13. e. 2.) That is, DD+ 2 Ba BB + CC. Demon Corollary. Hence it follows, that, if the Sides of any Acute-angled Triangle be known, the Perpendicular P, and the Segments of the Side whereon it falls (viz. a, e.) may be easily found. CHA P. IV. The Solution of feveral Easy Problems in Plain Geometry, whereby the Learner may (in Part) perceive the Application or Ufe of the foregoing Theorems. N OT E, when a Line, or the Side of any plain Triangle, is ang Way cut into two (or more Parts, either by a Perpendicular Line let fall upon it, or otherwife, thofe Parts are usually call'd Segments; and fo much as one of thofe Parts is longer than the other, is call'd the Difference of the Segments. And when any Side of a Triangle, or any Segment of its Side is given, 'tis ufually mark'd with a fmall Line cross it, thus: —|and thofe Sides, or Parts of Sides, that are fought, are mark'd with four Points, thus: :: PROBLEM I. To cut or divide a given Right-line (as S) into Extream and That is, to divide a Line fo, that the Square of the greater Segment (or Part) a, may be equal to the Rectangle made of the whole Line S, and the leffer Segment e. Viz. 1 Seaa, by the Problem. And 2 S S a ae, for Sati e 1-8 6, folved =S-a. By Axiom 5. 5 aa=SS_Sa 7a= √ SS + SS SS. See Pages 195, 196. Note, The laft Problem cannot be truly anfwer'd by Numbers, but Geometrically it may be performed, thus: 1. Make a Square, whofe Side is S the given Line, and bifect one of its Sides in the Middle, as at C; upon the Point C defcribe fuch a Semicircle as will pafs through the remoteft Points of the Square, and compleat its Diameter. S a 2. Then will either Part of the Diameter, on each End of the Side S, bea, the greater Segment fought. But a+S:S:S: a. By Theorem 13. Ergo, a a+Sa=SS. Which was to be done. PROBLEM II. The Bafe of any Right-angled Triangle, and the Difference between the Hypothenufe and Cathetus being given, to find the Cathetus, &c. 5 456 Let { And Then 4 a a b b + a a = d d + 2da + äa 5 bbdd2da dd 6 2da = bb 62d 7 a= 8 bb - - - dd 2d dd =65 Or, 8bd+2a:: db. By Theorem 13. 9 bbdd2da. As before at the 5th Step. T t Here Here you fee that either Way raises the fame Equation; neither is their any conftant Method or Road to be obferv'd in folving Geometrical Problems, but every one makes Use of such Ways and Theorems as happen to come first into their Mind, the Refult being every Way the fame. PROBLEM III. The Difference between the Bale and Hypothenule of any Right-angled Triangle, and the Difference between the Cathetus and Hypothenufe being both given, to find the Triangle. 5 6 dd+2da+aa=yy 7xx+2xa+aa=ee 8 dd+2dx+2da+2xa+xx+aa=□ Hypothenuse. 6+79dd+2da+2xa+xx+2aa=yy+ee. The two laft Steps are equal, by Theorem 11. Confequently, if thofe Things that are equal in both be taken away, the Remainders will be equal. By Axiom 2. The Hypothenufe, and the Sum of the other two Sides, of ang Rightangled Triangle, being given, thence to find the Sides. |