To a given right line to apply a parallelogram equal to a given rectilinear figure, and deficient by a figure similar to a given parallelogram. But the rectilinear figure must not be greater than the parallelogram applied to half the given line, whose defect is similar to the given parallelo gram. Statement.—Let AB be the given right line, and Z the rectilinear figure to which the parallelogram ap plied to AB is to be equal, not being greater than the parallelogram applied to the half line, with similar defect; and let X be the parallelogram to which the defect is to be similar. It is required to apply to AB a parallelogram which shall be equal to the given figure Z, and deficient by a parallelogram similar to X. Construction.-Bisect AB in E; describe upon AE a parallelogram AG similar to the given X; and complete APFB. AG is either equal to or greater than the given rectilinear figure Z. If it be equal, the problem is done. If it be greater, construct a parallelogram KLMN equal to its excess above Z, and similar to X (Prop. XXV.); since this parallelogram is less than AG, it is less than EF, which is equal to AG (I. 36); but it is similar to it, and therefore its sides KL and LM are less than the homologous sides EG and GF of the parallelogram EF; take away from these GK and GC, equal to KL and LM, and complete the parallelogram KGCD; this is similar to EF, since both are similar to X, and it is also similarly posited, therefore KGCD and EF are H about the same diagonal (Prop. XXVI.); draw their diagonal GDB, produce CD to S, and KD to M and N. Proof. Since the parallelogram EF is equal to the sum of KLMN and Z, but KC is equal to KLMN, the gnomon ENC is equal to Z; but ED and DF are equal (I. 43); therefore, if SN be added to both, EN and SF are equal; but since AE and EB are equal, EN is equal to ME (I. 36); therefore ME and SF are equal, and therefore, if ED be added to both, MS is equal to the gnomon ENC; but ENC is equal to the given rectilinear figure Z, therefore MS is equal to the given Z, and its defect SN, since it is similar to the parallelogram EF (Prop. XXIV.), is also similar to X (Prop. XXI.); and therefore, &c. Q. E. F. Annotation. It may be shown in a manner similar to the proof in the note to Prop. XXVII., that this Proposition is equivalent to the following Problem: To inscribe in a given triangle, a parallelogram equal to a given figure not greater than the maximum inscribed parallelogram, and having an angle in common with the triangle. PROPOSITION XXIX.-PROBLEM. Πρότασις κθ'. Παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι εὐθυγράμμῳ ἴσον παραλληλόγραμμον παραβαλεῖν ὑπερβάλλον εἴδει παραλληλογράμμῳ ὁμοίῳ τῷ δοθέντι. To apply to a given right line a parallelogram equal to a given rectilinear figure, and exceeding by a parallelogram similar to a given one. Statement. Let AB be the given line, and Z the given figure to which the applied parallelogram is to be equal, and let X be the parallelogram to which the excess is to be similar; it is required to apply to AB a Construction.—Bisect AB in E; upon EB construct a parallelogram EFLB, similar to the given parallelogram X, and construct a parallelogram GH similar to the parallelogram EL, and equal to the sum of EL and Z; since GH is greater than EL, its sides GK and KH are greater than the homologous sides of EL, which are FE and FL (Prop. XXII.); on these sides produced take FN and FM equal to GK and KH, and complete the parallelogram NM: this is similar to GH, therefore similar to EL, and it is similarly posited, and therefore they are about the same diagonal; draw the diagonal FBX, through A draw AC parallel to EN, until it meet PN produced.. Proof. Since NM and GH are equal, and GH is equal to the sum of Z and EL, NM is also equal to the sum of Z and EL; take away from both EL, and the gnomon NOL is equal to Z; but since AE and EB are equal, the parallelograms AN and EP are equal (I. 36), and also EP and BM are equal (I. 43); therefore AN is equal to BM; add ON to both, and AX is equal to the gnomon NOL, and therefore equal to the given rectilinear figure Z; and its excess PO is similar to the parallelogram EL, and therefore similar to the given figure X. Therefore, &c. Q. E. F. Annotation. This Proposition is equivalent to the following Problem : To exscribe to a given triangle a parallelogram equal to a given rectilinear figure, and having an angle equal to one of the angles of the given triangle. In the accompanying figure, ABC is the given triangle; AX is the inscribed, and AY the exscribed, parallelogram. A B PROPOSITION XXX.-PROBLEM. Πρότασις λ'.—Τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην ἄκρον καὶ μέσον λόγον τεμεῖν. To cut a given finite right line in extreme and mean ratio. Statement.-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 produce CA to G, so that AD described on the part produced may be a square similar to BC, and CD a rectangle equal to BC (Prop. XXIX.). A F E B Proof. Because BC is equal to CD, and the part CE common to both. From each take CE, and the remainder BF is equal to the remainder AD. Because BF and AD are equiangular and equal; therefore their sides about the equal angles at E are reciprocally proportional (Prop. XXIV.). Wherefore FE is to ED as AE is to EB. But FE is equal to AC (I. 34), that is, to AB; and ED is equal to AE. Therefore BA is to AZ 8 12 4 % 22 OTHERWISE: Statement. Let AB be the great is required to cut it in extreme and mean mis Construction. Divide 13 at point C, so that the rectangle tained by AB and BC may be 21 to A the square of AC IL 11. Proof.—Because the rectangle 130 la epal to the square of AC. Therefore BATOM 402. (Prop. XVII). Wherefore 13 is mt in extreme and mean ratio at C. Q. E. F. PROPOSITION XXXL-THEOREM. Πρότασις λα'. - Εν τοῖς ὀρθογωνίους τριγώνης, τα ἀπό τῆς τὴν ὀρθὴν γωνιάν υποτεινούσης πλευράς είλης ἴσον ἐστὶ τοῖς ἀπὸ τὴν ὀρθὴν γωνιάν περιέχουσών πλευρών είλεσι τοῖς ὁμοίους τε καὶ ὁμοίως ἀναγραφομένους, In right-angled triangles the 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. Statement.-Let ABC be a right-angled triangle, having the right angle BAC. The rectilinear figure described upon BC is equal to the similar and similarly described figures upon BA and AC. Construction.-Draw the perpendicular AD (I. 12), Proof. Because in the right-angled triangle ABC, AD is drawn from the right angle at A perpendicular to |