PROP. XVI. THEOR, If four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means: and if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines are proportionals. First, let the four straight lines AB, CD, E, F, be proportionals, viz. let AB be to CD as E to F: the rectangle contained by AB and F shall be equal to the rectangle contained by CD and E. E F H From the points A, C, draw AG, CH at right angles to AB, CD; make AG equal to F, and CH equal to E; and complete the parallelograms BG, DH: Then, because AB is to CD as E to F, and that E is equal to CH and F to AG, therefore AB is to CD as CH to AG, that is, the sides of the parallelograms BG, DH about the equal angles, are reciprocally proportional, and therefore (6. 14) the parallelogram BG is equal to the parallelogram DH: But the parallelogram BG is contained by the straight lines AB and F, because AG is equal to F, and the parallelogram DH is contained by CD and E, because CH is equal to E; therefore the rectangle contained by AB and F is equal to the rectangle contained by CD and E. B C D Next, let the rectangle contained by AB and F be equal to the rectangle contained by CD and E: then these four lines shall be proportionals, viz. AB shall be to CD as E to F. For, the same construction being made, because the rectangle AB, F, is equal to the rectangle CD, E, and that the rectangle BG is contained by AB and F, because AG is equal to F, and the rectangle DH by CD and E, because CH is equal to E, therefore the parallelogram BG is equal to the parallelogram DH: And they are equiangular: But (6. 14) the sides about the equal angles of equal parallelograms are reciprocally proportional; therefore, AB is to CD as CH to AG: And CH is equal to E, and AG to F; therefore AB is to CD as E to F. Wherefore, If four straight lines &c. Q.E.D. If three straight lines be proportionals, the rectangle contained by the extremes is equal to the square of the mean: and, if the rectangle contained by the extremes be equal to the square of the mean, the three straight lines are proportionals. First, let the three straight lines A, B, C be proportionals, viz. A to B as B to C: the rectangle contained by A and C shall be equal to the square of B. Take D equal to B: Then, because A is to B as B to C, and that B is equal to D, therefore A is to B as D to C: But, if four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means; A therefore the rectangle A, C, is B D B Next, let the rectangle contained by A, C be equal to the square of B: A shall be to B as B to C. For, the same construction being made, because the rectangle A, C is equal to the square of B, and that the square of B is equal to the rectangle B, D, because B is equal to D, therefore the rectangle A, C is equal to the rectangle B, D: But (6. 16) if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportionals; therefore A is to B as D to C: But B is equal to D; therefore A is to B as B to C. Wherefore, If three straight lines &c. PROP. XVIII. PROB. Q. E. D. Upon a given straight line to describe a rectilineal figure, similar and similarly situated to a given rectilineal figure. Let AB be the given straight line, and CDEF the given rectilineal figure of four sides: it is required to describe upon the given straight line AB, a rectilineal figure, similar and similarly situated to CDEF. H F K Join DF, and at the points A, B, in the straight line AB, make the angle BAG equal to the angle DCF, and the angle ABG to the angle CDF; therefore also the remaining angle CFD is equal to the remaining angle AGB, and the triangle CFD is equiangular to the triangle AGB: Again, at the points B, G, in the straight line BG, make the angle GBH equal to the angle DFE, and the angle BGH to the angle DFE; therefore the remaining angle DEF is equal to the remaining angle BHG, and the triangle DEF is equiangular to the triangle BHG: Then, because the angle AGB is equal to the angle CFD, and the angle BGH to the angle DFE, therefore the whole angle AGH is equal to the A B C D T whole angle CFE: In like manner, the angle ABH is equal to the angle CDE: And the angle BAG is equal to the angle DCF, and the angle BHG to DEF: Therefore the rectilineal figure ABHG is equiangular to CDEF: Also these figures have their sides about the equal angles proportionals: For, because the triangles BAG, DCF are equiangular, therefore BA is to AG as DC to CF (6.4): Again, for the like reason, AG is to GB as CF to FD, and GB to GH as FD to FE; therefore, ex æquali, AG is to GH as CF to FE: In like manner, it may be proved that AB is to BH as CD to DE: And GH is to HB as FE to ED: Therefore the figures ABHG, CDEF are equiangular, and have their sides about the equal angles proportionals: Wherefore (6. Def. 1) they are similar to one another. Next, let it be required to describe, upon the given straight line AB, a rectilineal figure similar, and similarly situated, to the rectilineal figure CDKEF. Join DE, and upon the given straight line AB describe, as in the former case, the rectilineal figure ABHG, similar, and similarly situated, to the quadrilateral figure CDEF; and at the points B, H, in the straight line BH, make the angle HBL equal to the angle EDK and the angle BHL equal to the angle DEK-therefore also the remaining angle at L will be equal to the remaining angle at K: Then, because the figures ABHG, CDEF are similar, the angle BHG is equal to the angle DEF: But the angle BHL is equal to the angle DEK; therefore the whole angle GHL is equal to the whole angle FEK: Similarly, the angle ABL is equal to the angle CDK: Therefore the fivesided figures ABLHG, CDKEF are equiangular: Also because the figures ABHG, CDEF are similar, therefore AB is to BH as CD to DE: But (6. 4) BH is to BL as DE to DK; therefore, ex equali, AB is to BL as CD to DK: In like manner, GH is to HL as FE to EK: and (6.4) BL is to LH as DK to KE: Therefore the figures ABLHG, CDKEF are equiangular, and have their sides about the equal angles proportionals: Wherefore they are similar to one another: And, in like manner, upon a given straight line may be described a rectilineal figure, similar, and similarly situated, to a given rectilineal figure of six sides; &c. PROP. XIX. THEOR. Q. E. F. Similar triangles are to one another in the duplicate ratio of their homologous sides. Let ABC, DEF be similar triangles, having the angle at B equal to the angle at E, and let AB be to BC as DE to EF, so that the side BC is homologous to EF: the triangle ABC shall be to the triangle DEF in the duplicate ratio of BC to EF. A D ÂÅ G F Take BG a third proportional to BC, EF (6. 11), so that BC may be to EF as EF to BG, and join AG: Then, because AB is to BC as DE to EF, therefore, alternately (5. 16), AB is to DE as BC to EF: But BC is to EF as EF to BG; therefore AB is to DE as EF to BG, that is, the sides of the triangles ABG, DEF, about their equal angles, are reciprocally proportional, and therefore (6. 15) the triangle ABG is equal to the triangle DEF: And because BC is to EF as EF to BG, therefore (5. Def. 10) BC is to BG in the duplicate ratio of BC to EF: But, as BC to BG, so is the triangle ABC to the triangle ABG; therefore the triangle ABC is to the triangle ABG in the duplicate ratio of |
