Hence (and from 34. 1.) it is manifeft, (1.) that Triangles having one Angle (at C) equal, have a Ratio to each other compounded of the Ratio's of the Right Lines AC to CB, and EC to CF, containing the Equal Angle. (2.) It is manifeft that Rectangles, and *confequently any Parallelograms, are to each other, in a Ratio, compounded of the Ratio's of the Bafe of the one to the Bafe of the other, and of the Altitude of the one, to the Altitude of the other. And you may reafon after the fame manner about 35. I. Triangles. L 4 (3.) It d 29. I. (3.) It is manifeft how to exprefs the Proportion of Triangles, and Parallelograms. Let there be two Parallelograms X & Z, whose Bases are AC CB, and Altitudes CL, CF. Make CL: CF:: CB: Q; then shall * X: Z:: AC: O. In every Parallelogram ABCD, the ParalleloE grams EG, HF, that are about the Diameter AC are fimilar to each other, and to the whole Parallelogram ABCD. For the Parallelograms EG, HF, have each one Angle common with the whole one. Therefore they are a mutually Equiangular to each other, and to the whole Parallelogram. Alfo the Triangles ABC, AEI, IHC; as likewife ADC, AGI, IFC, are Equiangular. Therefore AE: EI:: AB: BC. but AE: AI ::AB: AC; & AI: AG:: AC: AD. Whence by Equality, AE: AG:: AB: AD. Therefore 1 def. 6. the Pgr. EG, BD are fimilar. In like manner, HF, BD are fimilar. Therefore, &c. 4. 6. 22. 5. с b d PROP. XXV. To defcribe a right-lined Figure P fimilar and fimilarly fituate to a given right-lined Figure ABEDC, and equal to another given right-lined Figure F. Make the Rectang. AL ABEDC. also upon BL, make the Rectang. BM F. and between AB BH, find a mean Proportional NO; upon which make the Polygon P fimilar to the given Square ABEDC, and this fhall be equal to d the the given Figure F. for ABEDC (AL): P:: AB: BH:: AL': BM. Therefore PBM he cor. 20. If you deny that AC is the common Diameter, let it be the Diameter AHC cutting EF in H, and draw HI parallel to AE. The Parallelograms EI, DB are fimilar. Therefore AE: a 24. 6. ĚH :: AD: DC :: AE: EF. and fod EH = FE; which is absurd. с EH= b 1 def. 6. c byp. d 9. 5. PROP. 9 ax. ५ C one D or EG. Make FELIH; and FGM =IK, and thro' L, M draw the Parallels RN, MN; and AR parallel to NM. Also continue `out ABP, GBO, and draw the Diameter FBN: then is the Parallelogram AN that which is fought. C For the Parallelograms D, HK, LM, EG are fimilar. Therefore the Pgr. OP is fimilar to the Parallelogram LM or D. Alfo LM = HK EG+C. therefore C=Gnom. ENG. But ALLB BM; whence " CAN. Q. E. F. d To cut a given terminate right Line AB H into mean and extreme Proportion. (AB :AG:: AG: GB.) i Cut AB in G, so that AB x BG= AG' then is * BA: AG:: AG: GB. Q. E. F. PROP. XXXI k Any Figure BF defcribed upon the Side BC of a right-angled Triangle Subtending the right Angle BAC, is equal to the Figures BG, AL, described upon the Sides BA, AC, containing the right Angle, being fimilar and alike fituate to the former Figure. Let fall the Perpendicular AD from the right 1 cor. 8.6. Angle BAC; then because DC: CA :: CA: CB; therefore is " AL: BF:: DC: CB. Alfo fince DB: BA:: BA: BC, therefore is "BG: m cor.20.6. m BF E a BF::DB: BC. Whence AL + BG: BF :: COROL From this Prop. you may add and substract any fimilar Figures, after the fame manner as Squares are added and substracted in the Schol. of 47. I. PROP. XXXII. If two Triangles ABC, DCE, having two Sides proportional (AB: AC:: DC: DE) be fo com pounded or fet together at one Angle ACD, that their Homologous Sides are likewife Parallel (AB to DC, and |