a e d PROP. XXXVII, THEOR. Triangles (BAC and BDC) on the same base, and be- Fig. 56. tween the same parallels, are equal. Draw through the point C the right lines CE and CF parallel to BA and BD (1): the parallelo- (1) Prop.31. grams BAEC, BDFC are equal (2); but the triangles (3) Prop. 34. BAC and BDC are halves of them (3), and therefore (4) Ax. 7. equal (4). Prop.35. PROP. XXXVIII. THEOR. 1 Triangles (BAC and HDE) on equal bases, and between Fig. 57. the same parallels, are equal. Draw, through the points C and E, the right lines CF and EG parallel to BA and HD (1): the parallelo- (1) Prop.31. grams BAFC, HDGE are equal (2); but the triangles (2) Prop.36. BAC and HDE are halves of them (3), and therefore (3) Prop. 34. (4) 7. equal (4). PROP. XXXIX. THEOR. Equal triangles (BAC and BDC), on the same base Fig. 58. and on the same side of it, are between the same parallels. If the right line AD, which joins the vertices of the triangles, be not parallel to BC, draw through the point A a right line AF parallel to BC, cutting a side BD of the triangle BDC, or the side produced, in a point E different from the vertex; and draw CE. α Because the right lines AF and BC are parallel, the triangle BEC is equal to BAC (1); but BDC is also (1) Prop.37. equal to BAC (2); therefore BEC and BDC are equal (2) Hypoth. (3), a part equal to the whole, which is absurd. There- (3) Ax. 1. Fig. 60. fore the line AF is not parallel to BC: and in the same manner it can be demonstrated that no other line, except AD, is parallel to it; therefore AD is parallel to BC. PROP. XL. THEOR. Equal triangles (BAC and GDH), on equal bases and on the same side, are between the same parallels. If the right line AD, which joins the vertices of the two triangles, be not parallel to BH, draw through the point A the right line AF parallel to BH, cutting a side GD of the triangle GDH, or the side produced, in a point E different from the vertex; and join HE. Because the right line AF is parallel to BH, and BC and GH are equal, the triangle GEH is equal to BAC (1); but GDH is also equal to BAC (2); therefore GEH and GDH are equal (3), a part equal to the whole, which is absurd. Therefore AF is not parallel to BH: and in the same manner it can be demonstrated that no other line, except AD, is parallel to BH; therefore AD is parallel to BH. PROP. XLI. THEOR. If a parallelogram (BF) and a triangle (BAC) have the same base, and be between the same parallels, the parallelogram is double of the triangle. Draw CD. The triangle BDC is equal to the trian(1) Prop.37. gle BAC (1); but BF is double of the triangle BDC (2) Prop.34. (2); therefore BF is also double of the triangle BАС. Schol. Hence it is evident that a parallelogram is double of a triangle, if they have equal bases, and be between the same parallels. Fig. 61 ナ Cor. If a triangle BAC and a parallelogram DF be between the same parallels, and the base BC of the triangle be double the base DC of the parallelogram, the triangle is equal to the parallelogram. Draw AD. Because the triangles BAD and DAC are equal (1); BAC is double of the triangle DAC, but (1) Prop.38. the parallelogram DF is double of the same DAC (2); (2) Prop.41. therefore the triangle BAC and the parallelogram DF (3) Ax. 6. are equal (3). PROP. XLII. PROB. To construct a parallelogram equal to a given trian- Fig. 61. gle (BAC), and having an angle equal to a given one (0). a Through the point A draw the right line AF parallel to BC; bisect BC the base of the triangle in D; and at the point D, and with the right line CD, make the angle CDE equal to the given one O; through C draw CF! d parallel to DE, until it meet the line AF in F: DF is the required parallelogram. Because EF is parallel to DC (1), and CF parallel to (1) Constr. DE (1), DEFC is a parallelogram (2): it has also the (2) Def. 30. angle CDE equal to the given one O (1): and it is equal to the triangle BAC, because it is between the (3) Cor. same parallels and on half the base of the triangle (3). Prop. 41. PROP. XLIII. THEOR. 4 In a parallelogram (AC) the complements (AK and Fig. 62. KC), of the parallelograms (FH and GE) about the diagonal, are equal. Draw the diagonal DB, and, through any point in it K, draw the right lines FE and GH parallel to AB and BC: then FH and GE are the parallelograms about the diagonal, AK and KC their complements. f Because the triangles BAD and BCD are equal (1), (1) Prop.34. and the triangles BGK, KFD are equal to BEK, KHD (1), take away the equals BGK and KFD, BEK and KHD, from the equals BAD and BCD, and the remainders, namely the complements AK and KC, are equal (2). Cor. The parallelograms OF and EK about the diagonal of a square AD are squares. Because the triangle BAC is isosceles, and the angle (2) Ax. 3. (1) Cor. 3. at A right, ABC is half a right angle (1): since, then, in the triangle OBG, the angle at O is right, for it is (2) Prop. 29. external to a right angle (2), and the angle OBG half a right angle, the angle OGB must also be half a right (3) Prop. 32. angle (3); and therefore the sides OG and OB are equal (4) Prop. 6. (4); it is evident therefore, that OF is a square. In the same manner it can be demonstrated that EK is a square. Fig. 64. 小 A 4 PROP. XLIV. PROB. To a given right line (OS) to apply a parallelogram, which shall be equal to a given triangle, and shall 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 GO in R; upon RO construct a parallelogram RC equal to the given triangle, and having an 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. (2) Constr. (3) Prop. 43. Because ABDL is a parallelogram (2), and in it FS and RC are the complements of the parallelograms about the diagonal, FS is equal to RC (3), but RC is equal to the given triangle GHO (2); therefore FS is (4) Ax. 1. equal to GHO (4); and because the angle OFL is (5) Prop. 29. equal to the internal one BAF (5), and BAF equal to the external one BRO (5), OFL is equal to BRO; but BRO is equal to the given V (2), therefore OFL is equal to V. (6) Cor. Prop. 22. 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 (6): apply to the given line OS a parallelogram equal to GHO (7), and it shall be (7) Part. pr. equal to KNO; as is evident. PROP. XLV. PROВ. To construct a parallelogram equal to a given rectilineal figure (ABCED), and having an angle equal to a given one. e Resolve the given rectilineal figure into triangles. Construct a parallelogram RQ equal to the triangle BDA, and having an angle RIQ equal to the given H (1); on a side of it RV construct the parallelogram XV (1) Prop. 42. equal to the triangle CBD, and having an angle equal to the given one (2); and so on construct parallelograms (2) Prop. 44. 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. Because RV and IQ are parallel, the angle VRI, together with QIR, is equal to two right angles (3); but (3) Prop. 29. VRX is equal to QIR (4); therefore VRI with VRX (4) Constr. is equal to two right angles (5), and therefore IR and (5) Ax. 2. RX form one right line (6); in the same manner it can (6) Prop. 14. 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 VRI, 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 that VF and FY form one right line; therefore QY is a right line, and also it is parallel to IL (4): and because LY and RV are parallel to the same line XF, LY is parallel to RV (7); but (7) Prop. 30. IQ and RV are parallel, therefore LY is parallel to IQ (7), and therefore LIQY is a parallelogram (8): and (8) Def. 30. it has the angle LIQ equal to the given H, and is equal (9) Constr. to the given rectilineal figure ABCED (9). 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. |