Fig. 64. PROP. XLIV. PROB. To a given right line (OS) to apply a parallelogram, which shall be equal to a given triangle and have an angle equal to a given one (V). Let the given triangle be GHO, one of whose sides GO and the given line OS form one right line. Bisect GOin R, upon RO construct a parallelogram RC equal to the given triangle, and having the angle (1) Prop.42 BRO equal to the given one V (1); through S draw SD parallel to OC or RB until it meet BC produced to D; draw DO to meet BR produced to A, through A draw AL parallel to RS or to BD, and produce CO and DS to F and L. The parallelogram FS is applied to the given line OS, is equal to the given triangle GHO, and has the angle OFL equal to the given V. Because ABDL is a parallelogram (2), and in it FS and RC are the complements of the parallelo(3) Prop.45 grams about the diagonal, FS is equal to RC (3), but (4) Constr. RC is equal to the given triangle GHO (4), therefore (5) Ar. 1. FS is equal to GHO (5); and because the angle OFL (6)Prop.29 is equal to the internal one BAF (6), and BAF equal to the external one BRO (6), OFL is equal to BRO, (7) Constr. but BRO is equal to the given V (7), therefore OFL is equal to V. (2) Constr. (8) Cor. Prop. 22. (9) part. præc. If the given triangle be KNO, which has not a side forming one right line with the given, produce the given line OS until the produced part be equal to any side KO of the given triangle, and upon it construct a triangle GHO equal to KNO(8), apply to the given line OS a parallelogram equal to GHO (9), and it shall be equal to KNO, as is evident. PROP. XLV. PROB. To construct a parallelogram equal to a given recti- Fig. 65. lineal figure ABCED, and having an angle equal to a See N. given one. Resolve the given rectilineal figure into triangles. Construct a parallelogram RQ equal to the triangle BDA (1), and having an angel RIQ equal to the given (1) Prop.44 H, on a side of it RV construct the parallelogram XV equal to the triangle CBD, and having an angle equal to the given one (2), and so on construct parallelo (2) Prop.44 grams equal to all the triangles into which the figure is resolved, ILYQ is a parallelogram equal to the given rectilineal figure, and having an angle LIQ equal to the given one H, (6) Prop.14 Because RV and IQ are parallel, the angle VRI, together with QIR, is equal to two right angles (3), (3) Prop.29 but VRX is equal to QIR (4), therefore VRI with (4) Constr. VRX is equal to two right angles (5), and therefore IR (5) 4x 2. and RX form one right line (6); in the same manner it can be demonstrated that RX and XL form one right line, therefore IL is a right line; and because QV is parallel to IR, the angle QVR, together with VR1, is equal to two right angles (3), but IR is parallel to VF, and therefore IRV is equal to FVR (3), and therefore QVR with FVR is equal to two right angles, and QV and FV form one right line (6); in the same manner it can be demonstrated of VF and FY, therefore QY is a right line, and also is parallel to IL (7): and because (7) Constr. LY and RV are parallel to the same line XF, LY is parallel to RV (8), but IQ and RV are parallel, therefore LY is parallel to IQ (8), and therefore LIQY is a (8) Prop.30 parallelogram (9), and it has the angle LIQ equal to (9) Def. 30. the given H, and is equal to the given rectilineal figure (10) Constr. ABČED (10). Cor. 1. Hence a parallelogram can be applied to a given right line and in a given angle, equal to a given rectilineal figure, by applying to the given line a parallelogram equal to the first triangle. & Ax 2. Fig. 66. (1) Sch. Cor. 2. In the same manner a parallelogram can be applied to a given line, equal to the sum of two or more rectilineal figures. Cor. 3. A parallelogram can be constructed equal to the difference between two given rectilineal figures, by applying to the same right line at the same side of it, and in the same angle, parallelograms equal to the two figures, the difference between these parallelograms is a parallelogram, and equal to the difference of the given quantities. PROP. XLVI. PROB. On a given right line (AB) to describe a square. From either extremity of the given line AB draw a line AC perpendicular (1) and equal to it (2), through C draw CD parallel to AB (3), and through B draw BD parallel to AC; ABCD is the required square. (3) Prop 31 Because ABCD is a parallelogram (4) and the angle (4) Constr. (5) Cor. 1. A a right angle, the angles C, D and B are also right (5), Prop. 34. and because AC is equal to AB (4), and the sides CD (6) Prop.34 and DB are equal to AB and AC (6), the four sides AB, AC, CD, DB, are equal, therefore ABCD is a (7)Def. 31. square (7). Fig. 67. (1)hypoth.& Cor. 1. The squares of equal right lines AD and XS are equal. Draw the diagonals BD and YS. Because the right Def. 31. lines BA, AD are equal to YX, XS (1), and the angles A and X equal, the triangles BAD, YXS are equal (2)Prop. 4. (2), and therefore the squares AC and XZ which are (3) Prop.34 double of the triangles (3), are equal. Fig. 67. Cor. 2. If two squares AC and XZ be equal, their sides are equal. For, if possible, let one of them AD be the greater, (1) Prop. 3. take the line AF equal to XS (1), and AE equal to XY, and join EF. The triangle EAF is equal to YXS (2) Prop. 4, (2), but YXS is equal to BAD, as they are halves of the equal squares AC and XZ, therefore EAF is equal (3) Ax. 1, to BÁD (3), a part to the whole, which is absurd: therefore neither AD nor XS is greater than the other, therefore they are equal. PROP. XLVII. THEOR. In a right angled triangle (ABC) the square of the Fig. 61. side (AC) subtending the right angle, is equal to the Se N. sum of the squares of the sides (AB and CB) which contain the right angle. On the sides AB, AC, and BC describe the squares AX, AF and BI, draw BE parallel to either CF or AD, and join BF and AI. Because the angles ICB and ACF are equal (1), (1) Ar. 11. if BCA be added to both, the angles ICA and BCF are equal, and the sides IC, CA are equal to the sides BC, CF (2), therefore the triangles ICA (2) Def.51. and BCF are equal (3); but AZ is parallel to CI (5) Prop. 4. (4), therefore the parallelogram CZ is double of the (4) Prop.14 triangle ICA, as they are upon the same base CI, and between the same parallels (5); and the parallelogram CE is double of the triangle BCF, as they are upon the same base CF and between the same parallels (5); (5) Prop.41 therefore the parallelograms CZ and CE being double of the equal triangles ICA and BCF are equal to one another (6); in the same manner it can be demon- (6) Ar. 6. strated that AX and AE are equal; therefore the whole DACF is equal to the sum of CZ and AX. Cor. 1. Given in numbers any two sides of a right angled triangle, the third side can be found, for it is the square root of the sum or difference of the squares of the given lines, according as the given sides contain the right angle or not. Cor. 2. To find a square equal to the sum of two Fig. 69. or more given squares. Let the right lines A, B and C be sides of the given squares, construct a right angle FDE, and take on the sides of it the lines DG and DH equal to the given lines A and B (1), draw GH and take DI and DK (1) Prop. 5. equal to GH and to the given line C, the square of IK is equal to the squares of A, B and C, for it is equal to the squares of D1 and DK (2), but the square of DI (2)Prop.47 is equal to the squares of DG and DH (2), therefore the square of IK is equal to the squares of DK, DH and DG, or of the lines A, B and C, which are equal to them. Cor. 3. To find a line whose square is equal to the Fig. 70. difference between the squares of two given lines AB and BC. Produce either of the given lines AB till the part (1) Prop. 3. produced is equal to the other (1), from B as a centre and with the radius AB describe a semicircle AEF; through D draw DE perpendicular to AB and it is the required line, For draw BE. The square of BE, or of BA which is equal to BE, is equal to the squares of BD and (2) Prop.47 DE (2), therefore if the square of BD be taken away from the square of BA, the remainder is equal to the square of DE. Fig. 71. Cor. 4. If from any angle of a triangle ABC a perpendicular be let fall upon the opposite side, the difference between the squares of the sides AB and BC which contain that angle, shall be equal to the difference between the squares of the segments AD and DC of the side on which the perpendicular falls. For the square of the side AB is equal to the squares of AD and DB (1), and the square of BC is equal to (1) Prop.47 the squares of BD and DC (1), therefore the difference between the squares of AB and BC is equal to the difference between the sum of the squares of AD and DB and the sum of the squares of CD and DB (2), or taking away the common square of DB, equal to the difference between the squares of AD and CD. (2) Ax. 3. Fig 71. Cor. 5. The excess of the squares of either side above the square of the conterminous segment of the third side is the same. For it is evident that the ex(1) Prop.47 cess is the square of the perpendicular BD (1). Fig. 72. (1) Schol. PROP. XLVIII. THEOR. If the square of one side (AC) of a triangle (ABC) be equal to the sum of the squares of the other two sides (AB and BC), the angle (ABC) opposite to that side is a right angle. From the point B draw BD perpendicular (1) to one Prop. 11. of the sides AB which contain the right angle, and (2) Prop. 3. equal to the other BC (2), and join AD. (3) Prop. The square of AD is equal to the squares of AB and 47. BD (3), or to the squares of AB and BC, which is |