greater than the whole AF. Therefore, of all parallelograms applied, &c. Q. E. D. See N. † 27.6. * 10. 1. 18. 6. † 36. 1. * 25.6. † Constr. 21.6. PROP. XXVIII. PROB. To a given straight line to apply a parallelogram equal to a given rectilineal figure, and deficient by a parallelogram similar to a given parallelogram: but the given rectilineal figure to which the parallelogram to be applied is to be equal, must not be greater+ than the parallelogram applied to half of the given line, having its defect similar to the defect of that which is to be applied; that is, to the given parallelogram. Let AB be the given straight line, and C the given rectilineal figure to which the parallelogram to be applied is required to be equal, which figure must not be greater than the parallelogram applied to the half of the line, having its defect from that upon the whole line similar to the defect of that which is to be applied; and let D be the parallelogram to which this defect is required to be similar. It is required to apply a parallelogram to the straight line AB, which shall be equal to the figure C, H and be deficient from the parallelogram upon the whole line by a parallelogram similar to D. * * Τ G OF R A E SB L M C Divide AB into two equal parts in the point E, and upon EB describe the parallelogram EBFG similar and similarly situated to D, and complete the parallelogram AG, which must either be equal to C, or greater than it, by the determination. If AG be equal to C, then what was required is already done: for, upon the straight line AB the parallelogram AG is applied equal to the figure C, and deficient by the parallelogram EF similar to D. But, if AG be not equal to C, it is greater than it: and EF is equal + to AG; therefore EF also is greater than C. Make* the parallelogram KLMN equal to the excess of EF above C, and similar and similarly situated to D: then, since D is similar+ to EF, therefore also KM is similar to EF: let KL be the homologous side to EG, * 26.6. and LM to GF: and because EF is equal to C and KM together, EF is greater than KM; therefore the straight line EG is greater than KL, and GF than LM: make+ GX equal to LK, and GO equal to LM, † 3. 1. and complete the parallelogram XGOP: therefore +31. 1. XO is equal and similar to KM: but KM is similar to EF; wherefore also XO is similar to EF; and therefore* XO and EF are about the same diameter: let GPB be their diameter, and complete the scheme. Then, because EF is equal to C and KM together, and XO a part of the one is equal to KM a part of the other, the remainder, viz. the gnomon ERO is equal + +3 Ax. to the remainder C: and because OR is equal to XS, by adding SR to each, the whole OB is equal to the whole XB: but XB is equal to TE, because the base * 36.1. AE is equal to the base EB; wherefore also TE is equal to OB: add XS to each, then the whole TS is † 1 Ax. equal to the whole, viz. to the gnomon ERO: but it has been proved that the gnomon ERO is equal to C; and therefore also TS is equal to C. Wherefore the parallelogram TS, equal to the given rectilineal figure C, is applied to the given straight line AB deficient by the parallelogram SR, similar to the given one D, because SR is similar to EF. Which was to be done. * * PROP. XXIX. PROB. * 43. 1. 24. 6. To a given straight line to apply a parallelogram equal to See N. a given rectilineal figure, exceeding by a parallelogram similar to another given. Let AB be the given straight line, and C the given rectilineal figure to which the parallelogram to be applied is required to be equal, and D the parallelogram to which the excess of the one to be applied above that upon the given line is required to be similar. It is required to apply a parallelogram to the given straight line AB which shall be equal to the figure C, exceeding by a parallelogram similar to D. * 18. 6. * 25. 6. Divide AB into two equal parts+ in the point E, 10.1. and upon EB describe the parallelogram EL similar and similarly situated to D: and make* the parallelogram GH equal to EL and C together, and similar and similarly situated to D: wherefore GH is similar** 21. 6. to EL: let KH be the side homologous to FL, and KG to FE: and because the parallelogram GH is *26.6. *36.1. 43. 1. *24.6. *46.1. * 29.6. 14. 6. *34. 1. * 14. 5. K H greater than EL, therefore the side KH is greater than LM B MN is equal to EL and C: take away the common part EL; then the remainder, viz. the gnomon NOL, is equal to C. And because AE is equal to EB, the parallelogram AN is equal to the parallelogram NB, that is, to BM*: add NO to each; therefore the whole, viz. the parallelogram AX, is equal to the gnomon NOL: but the gnomon NOL is equal to C; therefore also AX is equal to C. Wherefore to the straight line AB there is applied the parallelogram AX equal to the given rectilineal figure C, exceeding by the parallelogram PO, which is similar to D, because PO is similar to EL. Which was to be done. PROP. XXX. PROB. To cut a given straight line in extreme and mean ratio. Let AB be the given straight line; it is required to cut it in extreme and mean ratio. Upon AB describe the square BC, and to AC* apply the parallelogram CD, equal to BC, exceeding by the figure AD similar to BC: then, since BC is a square, therefore also AD is a square: and because BC is equal to CD, by taking the common part CE from each, the remainder BF is equal to the remainder AD: and these figures are equiangular, therefore their sides about the equal angles are reciprocally proportional: therefore, as FE to ED, so AE to EB: but FE is equal * to AC, that is, to† AB; and ED is equal to AE; therefore as BA to AE, so is AE to EB: but AB is greater than AE; wherefore AE is greater than EB*: there fore the straight line AB is cut in extreme and mean ratio in E*. Which was to be done. Otherwise, Let AB be the given straight line; it is required to cut it in extreme and mean ratio. Divide AB in the point C, so that the rectangle contained by AB, BC may be A св equal to the square of AC: then, because 3 Def. 6. 11. 2. 17.6. the rectangle AB, BC, is equal to the square of AC; as BA to AC, so is AC to CB*: therefore AB is cut in extreme and mean ratio in C*. Which was to be done. * 3 Def. 6. In right angled triangles, the rectilineal figure described See N. upon the side opposite to the right angle, is equal to the similar and similarly described figures upon the sides containing the right angle. Let ABC be a right angled triangle, having the right angle BAC: the rectilineal figure described upon BC shall be equal to the similar and similarly described figures upon BA, AC. 8. 6. Draw the perpendicular+ AD: therefore, because † 12. 1. in the right angled triangle ABC, AD is drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC are similar to the whole triangle ABC, and to one another: and because the triangle ABC is similar to ADB, as CB to BA, so is* * 4. 6. BA to BD: and because these three straight lines are proportionals, as the first is to the third, so is the figure upon the first to the similar and similarly described figure* upon the second: therefore as CB to BD, so is the figure upon CB to the similar : B D C and similarly described figure upon * 2 Cor. 20. 6. See N. * 29. 1. † 1 Ax. * 6.6. † 2 Ax. * 32. 1. * 14. 1. See N. PROP. XXXII. THEOR. If two triangles which have two sides of the one proportional to two sides of the other, be joined at one angle so as to have their homologous sides parallel to one another; the remaining sides shall be in a straight line. Let ABC, DCE be two triangles which have the two sides BA, AC proportional to the two CD, DE, viz. BA to AC, as CD to DE; and let AB be parallel to DC, and AC to DE: BC and CE shall be in a straight line. D Because AB is parallel to DC, and A the straight line AC meets them, the alternate angles* BAC, ACD are equal; for the same reason, the angle CDE is equal to the angle ACD; wherefore also BAC is equal to CDE: and because the triangles ABC, DCE have one angle at A equal to one at D, and the sides about these angles proportionals, viz. BA to AC, as CD to DE, the triangle ABC is equiangular* to DCE: therefore the angle ABC is equal to the angle DCE: and the angle BAC was proved to be equal to ACD; therefore the whole angle ACE is equal † to the two angles ABC, BAC: add the common angle ACB, then the angles ACE, ACB are equal to the angles ABC, BAC, ACB: but ABC, BAC, ACB are equal to two right angles; therefore also the angles ACE, ACB are equal to two right angles and since at the point C, in the straight line AC, the two straight lines BC, CE, which are on the opposite sides of it, make the adjacent angles ACE, ACB equal to two right angles; therefore* BC and CE are in a straight line. Wherefore, if two triangles, &c. Q. E. D. PROP. XXXIII. * THEOR. In equal circles, angles, whether at the centres or circumferences, have the same ratio which the circumferences on which they stand have to one another: so also have the sectors. Let ABC, DEF be equal circles; and at their centres the angles BGC, EHF, and the angles BAC, EDF, |