འས (13) prop. 16. (10) prop. 4. and EFD are equal (10), but the angle C is also equal (11) hypoth to EFD (11), therefore AHB and C are equal (12), (12) 4.1. which is absurd (13); neither, therefore, of the sides BC and DF is greater than the other, therefore they are equal, but BA and DE are also equal, as also the (14) hypoth. angles B and D (14), therefore the side AC is equal to the side EF, and also the angle A to the angle (15) prop.4. E (15). Fig. 41. Schol. It is evident that the triangles themselves are equal. Cor. 1. In an isosceles triangle ABC the right line BD, drawn from the vertex perpendicular to the base, bisects the base and the vertical angle. For in the triangles ABD, CBD, the angles A and ADB are respectively equal to the angles C and CDB (1) hypotk. (1), and the side BD, which is opposite to the equal angles A and C, is common to both triangles, therefore the angles ABD and CBD are equal, and also the sides (2) prop. 26. AD and DC (2), and therefore the base and vertical angle are bisected. Fig. 42. Cor. 2. It is evident that the right line, which bisects the vertical angle of an isosceles triangle, bisects the base and is perpendicular to it; and that the right line drawn from the vertical angle, bisecting the base, is perpendicular to it, and bisects the vertical angle. PROP. XXVII. THEOR. If a line (EF) intersect two right lines (AB and CD) and make the alternate angles equal to each other (AEF to EFD), these right lines are parallel. For, if it be possible, let these lines not be parallel, but meetin G; the external angle A EF of the triangle EGF (1) prop. 18. is greater than the internal EFG (1), but it is also equal (2) hypoth. to it (2), which is absurd: therefore the lines A B and CD do not meet at the side BD, and in the same manner it can be demonstrated, that they do not meet at the side AC. Since, then, the right lines do not meet on either (3) Def. 28. side, they are parallel (3). A4 42 PROP. XXVIII. THEOR. 43 E A B Ifa right line (EF) intersect two rightlines (AB and Fig. 43. CD) and make the external angle equal to the internal and opposite angle upon the same side of the line (EGA to GHC, and EGB to GHD); or make the internal angles at the same side (AGH and CHG, or BGH and DHG) equal to two right angles, the two right lines are parallel to one another. First, let the angles EGA and GHC be equal; and since the angle EGA is equal to BGH (1), the angles (1) prop. 15. GHC and BGH are equal, but they are the alternate angles, therefore the right lines AB and CD are parallel (2). In the same manner the proposition can be demonstrated, if the angles EGB and GHD were given equal. (2) Prop. 27. Next let the angles AGH and CHG taken together and CD are parallel (5). In the same manner the (5) prop. 27. PROP. XXIX. THEOR. Ifa right line (EF) intersect two parallel right lines Fig. 44. (AB and CD), it makes the alternate angles equal (AGH to GHD, and CHG to HGB); and the external angle equal to the internal and opposite upon the same side (EGA to GHC, and EGB to GHD); and also the two internal angles at the same side (AGH and CHG, BGH and DHG) together equal to two right angles. First the alternate angles AGH and GHD are equal. For, if it be possible, let one of them AGH be greater than the other, and adding the angle BGH to both, AGH and BGH together are greater than BGH and GHD; but AGH and BGH together are equal to (1)Prop. 13. two right angles (1), therefore BGH and GHD are less than two right angles, and therefore the lines A B (2) Ax. 12. and CD, if produced, would meet at the side BD (2), (3) hypoth. but they are parallel (3), and therefore cannot meet, which is absurd; therefore neither of the angles AGH and GHD is greater than the other, they are therefore equal. In the same manner it can be demonstrated that the angles BGH and GHC are equal. Secondly. The external angle EGB is equal to the internal GHD. For the angle EGB is equal to the (4) Prop. 15, angle AGH (4), and AGH is equal to the alternate angle GHD (5), therefore EGB is equal to GHD (6). In the same manner it can be demonstrated that EGA and GHC are equal. (5) Part. præc. (6) Part. Thirdly. The internal angles at the same side, BGH and GHD, together are equal to two right angles. For, since the alternate angles GHD and AGH are equal (6), if the angle BGH be added to both, BGH and GHD together are equal to BGH and AGH, and (7)Prop.13. therefore are equal to two right angles (7.) In the same manner it can be demonstrated that the angles AGH and GHC together are equal to two right angles. Fig. 45. PROP. XXX. THEOR. If two right lines (AB, CF) be parallel to the same right line (DN), they are parallel to each other. Let the right line GP intersect them, the external (1)Prop.29. angle GLB is equal to the internal LON (1), and also the angle LON is equal to OPF, (1) therefore GLB is (2) Ax. 1. equal to OPF (2), and therefore the right lines AB (3) Prop.28. and CF are parallel (3). L B PROP. XXXI. PROB. Through a given point (C), to draw a right line pa- Fig.46. rallel to a given right line (AB). In the line AB take any point F, join CF, and at the point C and with the right line CF make the an gle FCE equal to AFC (1), but at the opposite side (1) Prop. 23. of the line CF, the line DE is parallel to AB. For the right line FC intersecting the lines DE and AB, makes the alternate angles ECF and AFC equal, and therefore the lines are parallel (2). PROP. XXXII. THEOR. (2) Prop. 27. If any side (AB) of a triangle (ABC) be produced, Fig. 47. the external angle (FBC) is equal to the sum of the two internal and opposite angles (A and C): and the three internal angles of every triangle taken together, are equal to two right angles. Through B draw BE parallel to AC(1). (1) Prop. 31. The angle FBE is equal to the internal angle A (2), (2) Prop. 29. and the angle EBC is equal to the alternate C (2), therefore the whole external angle FBC is equal to the two internal angles A and C. The angle ABC with FBC is equal to two right an gles (3), but FBC is equal to the two angles A and (3) Prop. 13. C (4), therefore the angle ABC, together with the an- (4) Part. gles A and C is equal to two right angles. Cor. 1. Hence, in a triangle, if one angle be right, the other two are together equal to a right angle; and if one angle be equal to the other two, it is a right angle. Cor. 2. If two triangles have two angles equal to two, the remaining angles are also equal. Cor. 3. In a right angled isosceles triangle, each angle at the base is half a right angle. Cor. 4. Each angle of an equilateral triangle is equal to a third part of two right angles. præc. Fig. 48. Fig. 49. See N. Cor. 5. Hence can be derived a method of trisecting a right angle FAC; take any portion of the side AC, and construct upon it an equilateral triangle CBA, and bisect the angle CAB by the right line AE; since CAB is equal to a third part of two right angles, it must be equal to two-thirds of one right angle; therefore BAF is the third part of a right angle, and therefore equal to BAE and EAC. Cor. 6. All the internal angles of any rectilineal figure, ABCDE, together with four right angles, are equal to twice as many right angles as the figure has sides. Take any point F within the figure and draw the right lines FA, FB, FC, FD, and FE. There are formed as many triangles as the figure has sides, and, therefore, all their angles taken together are equal to (1)Prop.32. twice as many right angles as the figure has sides (1), but the angles at the point F are equal to four right angles, (2), and therefore the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. (2) Cor. 3. Fig. 50. Cor. 7. The external angles of any reetilineal figure, are together equal to four right angles. For each external angle, with the internal adjacent to it, is equal (1) Prop. 13. to two right angles (1), therefore all the external angles, with all the internal, are equal to twice as many right angles as the figure has sides; but the internal angles, together with four right angles, are equal to twice as many right angles as the figure has sides (2), take away from both the internal angles, and the external angles remain equal to four right angles. (2) Cor. præc. Fig.51. PROP. XXXIII. THEOR. Right lines (AC and BD) which join the adjacent extremities of two equal and parallel right lines (AB and CD), are themselves equal and parallel. Draw the diagonal AD, and in the triangles CDA (1) Hypoth. and BAD, the sides CD and BA are equal (1), AD is common to both triangles, and the angle CDA isequal (2) Prop.29. to the alternate BAD (2), therefore the lines AC and |