Let ABCD be a parallelogram, of which the diameter is AC; and EG, HK parallelograms about the diameter. Then the parallelograms EG, HK shall be similar both to the whole parallelogram ABCD, and to one another. A E D B K H C Demonstration. Because DC, GF are parallels, 1. The angle ADC is equal to the angle AGF (I. 29); for the same reason, because BC, EF are parallels, 2. The angle ABC is equal to the angle AEF; and each of the angles BCD, EFG is equal to the opposite angle DAB (I. 34), and therefore they are equal to one another; wherefore 3. The parallelograms ABCD, AEFG, are equiangular; and because the angle ABC is equal to the angle A EF, and the angle BAC common to the two triangles BAC, EAF, they are equiangular to one another; therefore 4. As AB to BC, so is AE to EF (VI. 4); and because the opposite sides of parallelograms are equal to one another (I. 34), AB is to AD, as AỀ to AG (V. 7) ; and DC′ to CB, as GF to FE; and also CD to DA, as FG to GA; therefore 5. The sides of the parallelograms ABCD, AEFG about the equal angles are proportionals; # and they are therefore similar to one another (VI. Def. 1). For the same reason, the parallelogram ABCD is similar to the parallelogram FHCK; wherefore 6. Each of the parallelograms GE, KH is similar to DB; but rectilineal figures which are similar to the same rectilineal figure, are also similar to one another (VI. 21); therefore 7. The parallelogram GE is similar to KH. Wherefore, parallelograms, &c. Q.E.D. PROPOSITION 25.-Problem. To describe a rectilineal figure which shall be similar to one, and equal to another given rectilineal figure. Let ABC be the given rectilineal figure, to which the figure to be described is required to be similar, and D that to which it must be equal. It is required to describe a rectilineal figure similar to ABC, and equal to D. Construction. Upon the straight line BC describe the parallelogram BE equal to the figure ABC (I. 45, Cor.); also upon CE describe the parallelogram CM equal to D (I. 45, Cor.), and having the angle FCE equal to the angle CBL; therefore BC and CF are in a straight line, as also LE and EM (I. 29 and I. 14); between BC and CF find a mean proportional GH (NI. 13), and upon GH describe the rectilineal figure KGH similar and similarly situated to the figure ABC (NI. 18). Then KGH shall be the rectilineal figure required. Demonstration. Because BC is to GH as GH to CF, and that if three straight lines be 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 (NI. 20, Cor. 2); therefore 1. As BC to CF, so is the rectilineal figure ABC to KGH; but as BC to CF, so is the parallelogram BE to the parallelogram EF (NI. 1); therefore H 2. As the rectilineal figure ABC is to KGH, so is the parallelogram BE to the parallelogram EF (N. 11); and the rectilineal figure ABC is equal to the parallelogram BE (constr.); therefore and it is similar to ABC. 3. The rectilineal figure KGH is equal to the parallelogram EF (N. 14); but EF is equal to the figure D (constr.); wherefore also 4. KGH is equal to D; Therefore 5. The rectilineal figure KGH has been described similar to the figure ABC, and equal to D. Q.E.F. PROPOSITION 26.-Theorem. If two similar parallelograms have a common angle, and be similarly situated; they are about the same diameter. Let the parallelograms ABCD, AEFG be similar and similarly situated, and have the angle DAB common. Then ABCD and AEFG shall be about the same diameter. G B Construction. For if not, let, if possible, the parallelogram BD have its diameter AHC in a different straight line from AF, the diameter of the parallelogram EG, and let GF meet AHC in H; and through H draw HK parallel to AD or BC. Demonstration. Since the parallelograms ABCD, AKHG are about the same diameter, they are similar to one another (NI. 24); wherefore 1. As DA to AB, so is GA to AK (NI. Def. 1); but because ABCD and AEFG are similar parallelograms (hyp.), as DA is to AB, so is GA to AE; therefore 2. As GA to AE, so GA to AK (N. 11) ; that is, GA has the same ratio to each of the straight lines AE, AK; and consequently 3. AK is equal to AE (N. 9), the less equal to the greater, which is impossible; therefore 4. ABCD and AKHG are not about the same diameter ; wherefore 5. ABCD and AEFG must be about the same diameter. Therefore, if two similar, &c QE.D. PROPOSITION 27.-Theorem. Of all parallelograms applied to the same straight line, and deficient by parallelograms, similar and similarly situated to that which is described upon the half of the line; that which is applied to the half, and is similar to its defect, is the greatest. Let AB be a straight line divided into two equal parts in C; and let the parallelogram AD be applied to the half AC, which is there fore deficient from the parallelogram upon the whole line AB by the parallelogram CE upon the other half CB; of all the parallelograms applied to any other parts of AB, and deficient by parallelograms that are similar and similarly situated to CE, AD shall be the greatest. Let AF be any parallelogram applied to AK, any other part of AB than the half, so as to be deficient from the parallelogram upon the whole line AB by the parallelogram KH similar and similarly situated to CE. therefore K B E Then AD shall be greater than AF. First. Let AK the base of AF, be greater than AC the half of AB. 2. CG is equal to KE (Ax. 1) ; to each of these equals add CF; then H Construction. Because CE is similar to the parallelogram HK (hyp.), they are about the same diameter (VI. 26); draw their diameter DB, and complete the scheme. Demonstration. Then, because the parallelogram CF is equal to FE (I. 43), add KH to both; therefore 1. The whole CH is equal to the whole KE; but CH is equal to CG (I. 36), because the base AC is equal to the base CB; therefore 3. The whole AF is equal to the gnomon CHL (Ax. 2) ; 4. CE, or the parallelogram AD is greater than the parallelogram AF Next. Let AK the base of AF be less than AC. Demonstration. The same construction being made, because BC is equal to CA, therefore to each of these add AL; then 2. DH is greater than LG; but DH is equal to DK (I. 43); therefore 3. DK is greater than LG; therefore the parallelogram DH is equal to the parallelogram DG (I. 36); wherefore 4. The whole AD is greater than the whole AF. Therefore, of all parallelograms applied, &c. Q.E.D. PROPOSITION 28.-Problem. 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 (VI. 27) 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, and be deficient from the parallelogram upon the whole line by a parallelogram similar to D. Construction. Divide AB into two equal parts in the point E (I. 10), and upon EB describe the parallelogram EBFG similar and similarly situated, to D (NI. 18), and complete the parallelogram |