Simple Proof of Euclid’s Fifth Postulate

Here is a simple and elegant, yet thorough, way to prove Euclid’s Fifth Postulate. It states: “That if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.” (I. P.5.)

To prove Postulate 5 from neutral geometry we first demonstrate that parallel lines can be constructed without assuming their existence:

  1. Pick any three points A, B, and C.
  2. Connect points AB and AC using straight lines. (I. P.1.)
  3. If one segment is longer than the other cut the longer one off equal to the shorter. Suppose segment AB is longer. Cut it off equal to segment AC at point D. (I.3.)
  4. Extend segments CA and DA to the opposite side of circle AC and cut each segment off at points F and E, respectively, so that segments AC, AD, AE, and AF are all equal. (P.2.) Points C, D, E, and F, will then all be on the circumference of circle AC centered at point A. (I. Def. 15.)
  5. Angle CAD is equal to angle EAF. (I.15.) Angle CAE is also equal to angle DAF.
  6. Bisect angles CAD and EAF. (I.9.)
  7. Bisect angles CAE and DAF. (I.9.)
  8. Connect points C, D, F, and E using straight lines. (Note that these connecting lines were already constructed in steps 6 and 7–they form the bases of the equilateral triangles used to bisect the four angles as per Proposition 9.)
  9. Locate point G at the intersection of segment CD and bisector line ab. Locate point H at the intersection of segment CE and bisector line cd. Locate point I at the intersection of segment EF and bisector line ab. Locate point J at the intersection of segment DF and bisector line cd.
  10. Segments AC and AE are equal. (Step 4.) Segment AH is common to triangles AHC and AHE. Angles HAC and HAE are equal because line Ac bisects angle CAE. (Step 7.) Therefore, triangles AHC and AHE are equal by the side-angle-side method. (I.4.)
  11. Segments AD and AF are equal. (Step 4.) Segment AJ is common to triangles AJD and AJF. Angles JAD and JAF are equal because line Ad bisects angle DAF. (Step 7.) Therefore, triangles AJD and AJF are equal by the side-angle-side method. (I.4.)
  12. Segments AC and AD are equal. (Step 4.) Segment AG is common to triangles AGC and AGD. Angles GAC and GAD are equal because line Aa bisects angle CAD. (Step 6.) Therefore, triangles AGC and AGD are equal by the side-angle-side method. (I.4.)
  13. Segments AE and AF are equal. (Step 4.) Segment AI is common to triangles AIE and AIF. Angles IAE and IAF are equal because line Ab bisects angle EAF. (Step 6.) Therefore, triangles AIE and AIF are equal by the side-angle-side method. (I.4.)
  14. Triangle CAE is equal to triangle DAF. This is so because segments AC, AD, AE, and AF are all equal and angles CAE and DAF are equal (steps 4 and 5). (I.4.)
  15. Likewise, triangle CAD is equal to triangle EAF. (See steps 4, 5, and 14.)
  16. Segment CH is equal to half the length of segment CE because segments HC and HE are equal. (See Step 10.) (C.N.3.) Likewise, segment FJ is equal to half the length of segment FD. But segments CE and FD are equal. (Step 14.) Therefore, segments CH and FJ are equal.
  17. Angle ACH is equal to angle AFJ. (See Step 14.) Segments AC and AF are equal. (Step 14.)
  18. Triangle AHC is therefore equal to triangle AJF. (I.4.) Both are also equal to triangles AHE and AJD.
  19. By similar reasoning, triangles AGC, AGD, AIE, and AIF are all equal.
  20. Lines CE and DF are parallel. (I.27.) Segments CD and EF are equal. (Step 15.) Moreover, segments CF and DE are equal. (Step 19.) This supports the conclusion that lines CE and DF will never meet. (See I. Def. 23.)
    • At this point, we are done! The opposite legs of rectangle CDFE (CE and DF, CD and EF) are parallel to the corresponding bisector line, and diagonals CF and DE always cross both opposite legs, regardless of the angle. Suppose angle EFD is right, and angle DEF is a small amount less than right. Angle ED crosses line DF at point D, which on the same side of line EF as angle DEF.
  21. But for those who are unconvinced, locate point K along line GA such that segment GK is equal to segment GA. Construct straight segments CK and DK.
  22. Triangle CGK is equal to triangle CGA because segment GK is equal to segment GA (step 21), segment CG is common, and angles CGA and CGK are right angles. (See step 6 and 12 and I.P.4 (all right angles are equal).) For the same reasons, triangles DGA and DGK are equal.
  23. Triangles CGA and DGA have already been proved equal. (Step 12.) Therefore, triangles CGK and DGK are also equal. (Step 22.)
  24. Segments CK and DK are radii of a circle equal to circle AC but centered on point K. Construct circle CK. Extend lines CK and DK to points L and M along circle CK opposite points D and C. (I. Def. 15 and P.2.)
  25. Construct straight segment LM.
  26. Angle LKM is equal to angle CKD. (I.15.) Segments KL and KM are radii of circle CK and therefore equal to segments CK and DK. (Step 24.) Therefore, triangle LKM is equal to triangle CKD. (Compare step 15.)
  27. Bisect angles CKL and DKM. (I.9.)
  28. Locate point N at the intersection of lines CL and gh. Locate point O at the intersection of lines DM and gh. Locate point P at the intersection of lines LM and IK.
  29. Triangles LPK and MPK are equal. This is so because segment KP is common, segments KL and KM are equal (steps 24 and 26), angle PKM is equal to angle GKC (I.15), angle PKL is equal to angle GKD (I.15), and angles GKC and GKD are equal. (Step 23.) Triangles LPK and MPK are thus equal by the side-angle-side method. (I.4.)
  30. Because angles LPK and MPK are equal, they must be right angles. (I.P.4.)
  31. Moreover, triangles LKM and CKD have already been proven equal. (Step 26.) Therefore, triangles CGK, DGK, LPK, and MPK are all equal by either the side-side-side or the side-angle-angle method. (I.7 and 26.)
  32. But triangles CGK and DGK are also equal to triangles CGA and DGA. (Step 22.) Triangles CGA and DGA are equal to triangles EIA and FIA. (Step 19.) Thus, triangles EIA, FIA, CGA, DGA, CGK, DGK, LPK, and MPK are all equal.
  33. Angles CAG and EAI are equal. (Step 32.) Angles CAH and EAH are also equal. (Step 10.) Thus, angle CAG plus angle CAH is equal to angle EAI plus angle EAH. (I. C.N.2; see I.13.) Therefore, angles GAH and IAH are equal to each other.
  34. By the reasoning in step 33, angles GAJ and IAJ are also equal.
  35. Angles CAG and DAG are equal. (Step 32.) Angles CAH and DAJ are also equal. (Step 18.) Thus, angle CAG plus angle CAH is equal to angle DAG plus angle DAJ. (I. C.N.2.) Therefore, angles GAH and GAJ are equal.
  36. By the reasoning in step 35, angles IAH and IAJ are also equal.
  37. Angles GAH, GAJ, IAH, and IAJ are all equal. Accordingly, they are all right angles. (I.4.) GAI is a straight line perpendicular to HAJ, which is also a straight line. (I.14.)
  38. Angles CKG and LKP are both equal to angle CAG. (Step 32.) These angles are all also equal to angles DAG, DKG, and MKP. (Id.)
  39. Segments CK and LK are equal. (Step 26.) Angles CKN and LKN are equal. (Step 27.) Segment KN is common. Thus, triangles CKN and LKN are equal. (I.4.) By the same reasoning, triangles DKO and MKO are equal.
  40. Angles CKG and LKP are equal. (Step 32.) Angles CKN and LKN are also equal. (Step 39.) Thus, angle CKG plus angle CKN is equal to angle LKP plus angle LKN. (I. C.N.2; see I.13.) Therefore, angles GKN and PKN are equal to each other.
  41. By the reasoning in step 40, angles GKO and PKO are also equal.
  42. Angles CKL and DKM are equal. (I.15.) Segments CK, DK, LK, and MK are all equal. (Step 26.) Triangles CKL and DKM are therefore equal. (I.4.)
  43. Line NO bisects angles CKL and DKM. (Step 27.) It follows that angles CKN, LKN, DKO, and MKO are all equal. (C.N.3.) From step 42, triangles CKN, LKN, DKO, and MKO are all equal. (I.26.)
  44. Angles CKG and DKG are equal. (Step 32.) Angles CKN and DKO are also equal. (Step 43.) Thus, angle CKG plus angle CKN is equal to angle DKG plus angle DKO. (I. C.N.2.) Therefore, angles GKN and GKO are equal.
  45. By the reasoning in step 44, angles PKN and PKO are also equal.
  46. Angles GKN, GKO, PKN, and PKO are all equal. Accordingly, they are all right angles. (I.4.) GKP is a straight line perpendicular to NKO, which is also a straight line. (I.14.) Moreover, point K is on straight line GAI (step 21), so points I, A, G, K, and P are all on a straight line.
  47. Straight segments DE and DL are equal radii of circle DE. (I. Def. 15.) Likewise, segments CF and CM are equal radii of circle CF. (Id.) Circles DE and CF are equal. (Step 32.)
  48. Points E, D, and L form an isosceles triangle. (I. Def. 20.) Points F, C, and M also form an isosceles triangle equal to EDL.
  49. Segment EL is a straight line. If point C lies on line EL, then segments CE and CL form a straight line. (I. Def. 4.) Suppose point C does not lie on straight line EL:
  50. Point C’ lies at the intersection of straight lines EL and DG. If segment CD is longer than segment, C’D, then angle CKD must be greater than angle LKM. (I.18.)
  51. But this is impossible, because angles CKD and LKM are equal. (Step 26.) Therefore, point C is the same as point C’. Line ECL is straight.
  52. By the same reasoning, line FDM is also straight.
  53. Angle CKN is equal to a right angle less angle CKG. (Steps 40 and 46.) Likewise, angle CAH is equal to a right angle less angle CAG. (Steps 33 and 37.) But angles CAG and CKG are equal. (Step 32.) Thus, angle CKN is equal to angle CAH. (C.N.3.)
  54. Just as line HJ bisects angle CAE (step 7), line NO bisects angle CKL. (Step 27.) From step 53, angles CAE and CKL are equal. (C.N. 2.)
  55. Triangles CAE and CKL are equal. (I.4; see steps 26 and 54.) Likewise, triangles DAF and DKM are equal.
  56. From step 55, angles CEA and CLK are equal. Yet these angles also form isosceles triangle EDL. (Step 48). Each side of triangle EDL is double the length of the corresponding sides of triangles CEA and CLK. This is considered equivalent to proving the parallel postulate.
  57. Furthermore, because angles NCK and GCK are equal to angles HCA and GCA, angles GCH and GCN are right angles. (I.P.4.) Angles GDJ and GDO are right angles for the same reason. Following the procedure set forth herein, angles IEH, IFJ, NLP, and OMP can also be shown to be right angles. Thus, the fact that segments EF and LM are equal to segment CD (step 32) demonstrates that lines EL and FM are equidistant as measured by three perpendiculars.
  58. Lines CE and DF are confirmed to be parallel because angles HCA and JFA are equal. (See step 20 and I.27–Prop. 27 is not being used here to prove lines CE and DF are parallel, but rather to confirm that what was herein proven meets Euclid’s definition of “parallel.”) Lines CE and DF will never meet because they are equidistant at three points. (See step 57 and I. Def. 23.) Any number of circles of radius AC can be constructed along line GAI to demonstrate this. Likewise, lines CD and EF are parallel.
    • At this point, the core of Euclid’s Fifth Postulate has been proven. Parallel lines have been constructed without resorting to assumptions other than those included in the first four postulates in Book I. However, to complete the proof, a few additional steps are necessary.
    • The Fifth Postulate involves three straight lines. One of them crosses the other two. Let lines CE and DF be the two straight lines. Does the angle at which the third line crosses them make a difference?
  59. Angles ACe plus angle ACH is equal to two right angles. (I.13.) (Proposition 13 demonstrates that supplementary angles add up to two right angles. Euclid does not use the term “supplementary,” but his use of “two right angles” signifies that intent.) Angle AFJ is equal to angle ACH. (Step 17.) Angles AFJ and AFf add up to two right angles. (I.13.) This will also be true of angles ADJ and its supplement. The same is true of angles AHC plus AHE and angles AJD and AJF. (I.13 and Def.10.)
  60. Therefore, angle ACe plus angle AFJ equals two right angles. (I. C.N. 2; see also I.28 and 29.) Likewise, angles AHC and AJD add up to two right angles.
    • This demonstrates that the angle at which a straight line crosses lines CE and DF will not matter for purposes of proving Euclid’s Fifth Postulate.
    • For simplicity’s sake, we will henceforth make the third (crossing) line perpendicular to one of the two other lines.
    • We must next prove that two lines whose interior angles form less than two right angles will meet on that side of the third (crossing) line.
  61. Let AB and AC be perpendicular lines:
  62. Let line CD intersect line AC at point C at an angle such that angle ACD is less than a right angle. Since angle CAB is right, angles ACD plus angle CAB total less than two right angles. (C.N.2.)
  63. Draw circle AB and locate point E at the intersection of circle AB and line CD.
  64. Locate point F at the intersection of circle AB and line AC and draw circle FE.
  65. Place point G at the intersection of circles AB and FE and join them to form line EG.
  66. Connect points A, E, F, and and G by straight line segments.
  67. Segments AE and AG are equal because they are both radii of circle AB. (I. Def. 15.) Segment AF (also a radius of circle AB) is common to triangles AEF and AGF. Segments FE and FG are equal because They are both radii of circle FE. (I. Def. 15.) Therefore, triangles AEF and AGF are equal by the side-side-side method. (I.8.)
  68. Locate point H at the intersection of lines AF and EG.
  69. Segments AE and AG have been shown to be equal. (Step 29.) Angles EAF and GAF have been shown to be equal. (Id.) Segment AH is common to triangles AHE and AHG. Therefore, triangles AHE and AHG are equal. (I.4.)
  70. If angles AHE and AHG are equal (see step 31), then they must be right angles. (I.P.4.) Angle CHE is supplementary to right angle AHE and is therefore a right angle equal to angle CAB, which is also a right angle. (Step 23.) But line CE crosses line HE. Therefore, if line CE is extended indefinitely, it will also cross line AB, which is parallel to line HE. (See Step 18.)

Q.E.D.

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