For, because AC bisects the parallelogram BD (34. 1), the triangles ABC, ADC are equal; and, because the same diagonal bisects the parallelograms EH, GF (34. 1), the triangles AEK, KGC are severally equal to the triangles AHK, KFC; therefore the triangles AEK, KGC together, are equal to AHK, KFC together (Ax. 2), which being taken from the equal triangles ABC, ADC, the remaining complements BK, KD are equal (Ax. 3). PROP. XLIV. PROB. To a given right line (AB), to apply a parallelogram, equal to a given triangle (C), and having an angle, equal to a given rectilineal angle (D). Make the parallelogram EFGH equal to C, having an angle EFG equal to D (42. 1); in EF produced, take Fi equal to AB (3. 1), complete the parallelogram GI (Cor. 6. 34. 1), and join FK; because the angle FGK is equal to the internal remote angle EHG (29. 1), and the angles FGK, FKG are together less than two right angles (17. 1), the angles EHK, FKH are also together less than two right angles, therefore HE, KF may be so produced towards E, F, as to meet (Theor. at 29. 1); let them, being produced, meet, as in 1, and through L, draw LN parallel to EI, meeting GF, KI produced in M and N; make the angle BAO equal to IFM (23. 1), AO to FM, and complete the parallelogram AOPB (Cor. 6. 34. 1) In the parallelogram HN, the complements HF, FN are equal (43. 1), but the parallelogram HF is equal to C (Constr.), therefore the parallelogram FN is equal to C (Ax. 1); and the parallelogram AP is equal to FN (Cor. 3. 34. 1), therefore the parallelogram AP is also equal to C (Ax. 1), and it is applied to the given right line AB, having an angle A, equal to the angle MFI (Constr.), and therefore [15. 1], to EFG, and therefore [Constr.], to D. PROP. XLV. PROB. To make a parallelogram, equal to a given right-lined figure (ABCD), and having an angle, equal to a given rectilineal angle (E). Join AC, and make the parallelogram KG, equal to the triangle ACD, having the angle K, equal to the angle E (42. 1); and to the right line GH, apply the parallelogram GM, equal to the triangle ABC, having the angle GHM equal to E (44. 1); the angles GHM and K, being each equal to the angle E (Constr.), are equal to each other, adding to each the angle GHK, the angles GHM, GHK together, are equal to K and GHK together (Ax. 2); but the angles K and GHK together, are equal to two right angles (29. 1), therefore the angles GHM, GHK together, are equal to two right angles, and therefore the right lines KH, HM make one right line (14. 1); and, since the angles HGF, HGL are severally equal to their alternate angles GHM, GHK (29. 1), the angles HGF, HGL together, are also equal to two right angles, and therefore the right lines FG, GL make one right line (14. 1); and FK and LM, being each of them parallel to GH, are parallel to each other (30. 1); and FL is parallel to KM; therefore FM is a parallelogram, equal to the given right-lined figure ABCD (Constr. and Ax. 2), and having an angle K equal to E (Constr.) 1 Cor.-Hence it appears, how there may be applied, to a given right line FK, a parallelogram, equal to a given rectilineal figure ABCD, and having an angle, equal to a given rectilineal one E; namely, by applying to FK, a parallelogram, equal to the triangle ACD, having an angle, equal to the given one (44. 1), and then proceeding, as in this proposition. PROP. XLVI. PROB. On a given right line (AB), to constitute a square. From the point A, draw AC at right c angles to AB (11. 1), on which take ÅD D equal to AB (3. 1), and complete the parallelogram AE (Cor. 6. 34. 1). Because AB, AD are equal (Constr.), and the sides DE, EB opposite to them, severally equal to AB, AD (34. 1), the four sides AB, AD, DE, EB are equal to each other, and the parallelogram AE is equilateral: it is also right angled, for the angle A being right (Constr.), its other angles are right (Cor. 1. 34. 1)); therefore ABED is a square (Def. 36), and is constituted on the given right line AB. A B Cor. 1.-Equal squares (ABCD, EFGH), have equal sides, (as AB, EF). For if AB and EF be not equal, let one of them, as AB, be the greater, and take thereon AI equal to EF, on which make the square AIKL (by this prop.); and because AI, AL are equal (Constr. and Def. 36), as also AB and AD (Hyp. and Def. 36), and AI is less than AB, AL is less than AD, and the square AIKL contained within the square ABCD; but the squares AIKL, EFGH, having equal sides, are equal (Cor. s. 34. 1), and the squares ABCD, EFGH are also equal (Hyp.) whence, the squares AIKL and ABCD, being each equal to EFGH, are equal to each other (Ax. 1), part and whole, which is absurd (Ax. 9); therefore AB is not greater than EF. In like manner, it may be shewn, that AB is not less than EF. Therefore AB, being neither greater nor less than EF, is equal to it. Cor. 2.-Parallelograms (CK, HF), about a diagonal (DB) of a square (AE), are squares. a H A B Because all the angles of the triangle ABD are equal to two right angles (32. 1), and the angle A a right angle (Hyp. and Def. 36), the angles ABD, ADB are together equal to a right angle, and being, because of the equal sides AB, AD, equal to each other (5. 1), either of them, as ABD, is half a right angle; and, in the triangle CBG, the angle BCG is equal to the internal remote on the same side A (29. 1), and therefore a right angle, and CBG has been shewn to be half a right angle, therefore CGB is also half a right angle; whence, the angles CBG, CGB being equal, CB is equal to CG (6, 1); and, in the parallelogram CK, the sides opposite to these are severally equal to them (34. 1); therefore the parallelogram CK is equilateral; and because its angle at B is right (Hyp. and Def. 36), right angled (Cor. 1. 34. 1); it is therefore a square (Def. 36) In like manner it may be demonstrated, that HF is a square. PROP. XLVII. THEOR. In every right angled triangle (ABC), the square of the side (BC), opposite the right angle (BAC), is equal to the squares of the other sides (AB and AC). On the sides BC, BA, AC, describe the squares BE, FA, AK (46. 1), through A, draw AL parallel to BD or CE (31. 1), F and join AD, FC. B Because the angles BAC, BAG are right angles (Hyp. and Def. 36), and therefore together equal to two right angles, the right lines GA, AC make one right line (14, 1); and because the angles FBA, DBC, being each of them right angles (Def. 36), are equal (Theor. at 11. 1), adding to each the angle ABC, the angles FBC, ABD are equal (Ax. 2), and the sides FB, BC are severally equal to the sides AB, BD (Def. 36), therefore the triangles FBC, ABD are equal (4. 1); but the parallelogram FA is double the tri D 1 angle FBC, being on the same base FB, and between the same parallels FB, GC (41. 1); and the parallelogram BL is double the triangle ABD, being on the same base BD, and between the same parallels BD, AL (41. 1); whence, the parallelograms FA, BL, being double the equal triangles FBC, ABD, are equal (Ax. 6). In like manner, joining BK, AE, the parallelograms AK, CL may be proved equal; therefore the square BE of BC is equal to the squares FA, AK of the other sides AB, AC (Ax. 2). Cor. 1.-Any number of squares being given, to find one equal to them all. Let the right lines A, B and C H be sides of the given squares; draw F D ABG DF, DK equal to C (3. 1); join KI, the square of which is equal to the squares of A, B and C ; for it is equal to the squares of DI, DK (47. 1), and the square of DI, or its equal HG, is equal to the squares of DG, DH (47. 1); therefore the square of KI is equal to the squares of DG, DH and DK, or of their equals A, B and C. Cor. 2. Two unequal right lines (AB and C) being given, to find a right line, whose square is equal to the excess of the square of the greater (AB), above that of the less (C). From the centre B, at the distance BA, describe the semicircle AFD (Post. 3), meeting AB produced in D; BD and BA are equal (Def. 10), and therefore BD greater than C; on BD take BE equal to Ĉ (3. 1), and from the point E, draw EF perpendicular to BD (11. 1), meeting the arch AFD in F; EF is the right line required. |