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; E O F R S B 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 (I. 36); 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 (VI. 25); then, since D is similar to EF (constr.), therefore also KM is similar to EF (VI. 21), let KL be the homologous side to EG, and LM to GF; and because EF is equal to Cand 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 (I. 3), and complete the parallelogram XGOP (I. 31); therefore X 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 (VI. 26); let GPB be their diameter, and complete the scheme. Then TS shall be the parallelogram required. M D Demonstration. 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, add XS to each, then 1. The remainder, viz., the gnomon ERO, is equal to the remainder C (Ax. 3) ; 3. TE is equal to OB (Ax. 1); and because OR is equal to XS, by adding SR to each (I. 43), 2. The whole OB is equal to the whole XB; but XB is equal to TE, because the base AE is equal to the base EB (I. 36); wherefore also 4. The whole TS is 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 5. 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 (VI. 24). Q.E.F. PROPOSITION 29.-Problem. To a given straight line to apply a parallelogram equal to a given rectilineal figure, exceeding by a parallelogram similar to another given. D 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. F K N L M P H C Construction. Divide AB into two equal parts in the point E (I. 10), and upon EB describe the parallelogram EL similar and similarly situated to D (VI. 18); and make the parallelogram GH equal to EL and C together, and similar and similarly situated to D (VI. 25); wherefore 1. GH is similar to EL (VI. 21); let KH be the side homologous to FL, and KG to FE; and because the parallelogram GH is greater than EL, therefore 2. The side KH is greater than FL, and KG than FE; produce FL and FE, and make FLM equal to KH, and FEN to KG, and complete the parallelogram MN; therefore 4. EL and MN are about the same diameter (VI. 26); draw their diameter FX, and complete the scheme. 3. MN is equal and similar to GH; but GH is similar to EL; wherefore MN is similar to EL; and consequently Then AX shall be the parallelogram required. Demonstration. Since GH is equal to EL and C together; and that GH is equal to MN; therefore F L M add NO to each; therefore X K 2. The remainder, viz., the gnomon NOL, is equal to C. And because AE is equal to EB, H 3. The parallelogram AN is equal to the parallelogram NB, (I. 36), that is, to BM (I. 43); 4. The whole, viz., the parallelogram AX, is equal to the gnomon NOL; but the gnomon NOL is equal to C; therefore also 5. AX is equal to C. Wherefore to the given 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 (VI. 24). Q.E.F. PROPOSITION 30.-Problem, 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. Construction. Upon AB describe the square BC (I. 46), and to AC apply the parallelogram CD, equal to BC, exceeding by the figure AD similar to BC (VI. 29). Then AB shall be cut in extreme and mean ratio in the point E. 1. AD is a square; and because BC is equal to CD, by taking the common part CE from each, 2. The remainder BF is equal to the remainder AD; and these figures are equiangular, therefore their sides about the equal angles are reciprocally proportional (VI. 14); therefore E F 3. As FE to ED, so AE to EB; but FE is equal to AC (I. 34), that is, to AB (Def. 30); and ED is equal to AE; therefore 4. As BA to AE, so is AE to EB; but AB is greater than AE; wherefore 5. AE is greater than EB (V. 14); therefore 6. The straight line AB is cut in extreme and mean ratio in the point E (VI. Def. 3). Q.E.F. 1 с A B Otherwise. Let AB be the given straight line. It is required to cut it in extreme and mean ratio. Construction. Divide AB in the point C, so that the rectangle contained by AB, BC, may be equal to the square on AC (II. 11). Then AB shall be cut in extreme and mean ratio in C. Demonstration. Because the rectangle AB, BC is equal to the square on AC; 1. As BA to AƠ, so is AC to CB (VI. 17) ; therefore 2. AB is cut in extreme and mean ratio in C (VI. Def. 3). Q.E.F. ROPOSITION 31.-Theorem. In right-angled triangles, the rectilineal figure described 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. Then the rectilineal figure described upon BC shall be equal to the similar and similarly described figures upon BA, AC. B Construction. Draw the perpendicular AD (I. 12). Demonstration. Because in the right-angled triangle ABC, AD is drawn from the right angle at A perpendicular to the base BC, 1. The triangles ABD, ADC are similar to the whole triangle ABC, and to one another (VI. 8); and because the triangle ABC is similar to ADB, 2. As CB to BA, so is BA to BD (VI. 4); and inversely, 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 (VI. 20, Cor. 2); therefore 3. As CB to BD, so is the figure upon CB to the similar and similarly described figure upon BA; 4. As DB to BC, so is the figure upon BA to that upon BC (V. B) ; for the same reason, as DC to CB, so is the figure upon CA to that upon CB; therefore 5. As BD and DC together to BC, so are the figures upon BA, AC to that upon BC (V. 24) ; but BD and DC together are equal to BC; therefore 6. The figure described on BC is equal to the similar and similarly described figures on BA, AC (V. A). Wherefore, in right-angled triangles, &c. Q.E.D. PROPOSITION 32.-Theorem. 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 |