Replacing Euclid’s Fifth Postulate

I’ve thought quite a bit about how to simplify the proof of Euclid’s Fifth Postulate. I think Euclid’s method of breaking up complicated proofs into short propositions that build upon another is a good one. Therefore, I propose to add the following propositions between Proposition 28 and 29 of Book I of Euclid’s Elements. (Proposition 29 is the first one in which he uses his Fifth Postulate.) Accordingly, they are numbered Propositions 28.1 through 28.13. Together, they prove the Fifth Postulate from neutral geometry, and render the assumption of parallel lines unnecessary.

Proposition 28.1.  The four intersections of two diameters with a circle form a quadrilateral.

  1. Take any given plane rectilineal angle.  Let point B be the intersection of the two legs of the angle.  Construct circle BA, and locate point C at the intersection of the other leg of the angle with the circle.  (I. Def. 8, 9, 15.)
  1. Produce segments BA and BC beyond point B through the opposite point on circle BA.  (P.2.)  Locate point D on straight line AB at the opposite intersection with circle BA.  Locate point E on straight line CB at the opposite intersection with circle BA.
  2. Line segments BA, BC, BD, and BE are all radii of circle BA.  (P.3.)
  3. Segments AD and CE are diameters of circle BA.  (I. Def. 17.)
  1. Connect points A and C, C and D, D and E, and E and A using straight lines.
  2. ACDE is a rectilineal figure contained by four straight lines and is therefore a quadrilateral.  (I. Def. 19.)

Proposition 28.2.  The opposite sides of a quadrilateral formed from the four intersections of two diameters with a circle are equal.

  1. Segments BC and BD are equal because they are both radii of circle BA.  (I.28.1.3.)
  2. Triangle CBD is isosceles.  (I. Def. 20.)
  3. Segments BA and BE are also equal to segment BC.  (I.28.1.3.)
  4. Triangle ABE is isosceles.
  5. Angle CBD is equal to angle ABE.  (I.15.)
  6. Triangle ABE is therefore equal to triangle CBD.  (I.4.)
  7. By the same reasoning, isosceles triangle DBE is equal to triangle ABC.
  8. Therefore, segments AE and CD are equal, and segments AC and DE are equal.
  9. Figure ACDE must be either a rhomboid or an oblong (rectangle).  (I. Def. 22.)

Proposition 28.3.  The straight lines bisecting the angles formed by the four points at the intersections of two diameters of a circle and the center of the circle are perpendicular.

  1. Bisect angle CBD.  (I.9.)
  1. Locate point G at the intersection of segment CD with bisector line Bb.
  2. Bisect angle ABE and locate point I opposite point G.
  3. Bisect angle ABC and locate point F at the intersection of segment AC with bisector line Ba.  Likewise, bisect angle DBE and locate point H opposite point F.
  4. Segment BG is common to triangles BGC and BGD.  Angles CBG and DBG are equal because line BG bisects angle CBD.  (C.N.3.)  Segments BC and BD are equal radii of circle BA.  (I.28.1.3.)  Therefore, triangles BGC and BGD are equal.  (I.4.)
  5. Following the reasoning in I.28.3.5, triangles BIA and BIE are also equal.
  1. Segment CD is a straight line.  (I.28.1.5.)  Angles BGC and BGD are equal, and they total two right angles.  Therefore, angles BGC and BGD are right angles.  (P.4 and I.13.)  For the same reason, angles BIA and BIE are also right.
  2. Segment AC is straight.  (I.28,1,5,)  Triangles BFA and BFC are equal.  (I.28.3.6.)  Likewise, triangles BHD and BHE are equal.  Thus, angles BFA, BFC, BHD, and BHE are all right angles.  (I.28.3.7.)
  3. Angles CBF and DBH are equal.  (I.28.3.8.)  Angles CBG and DBG are equal.  (I.28.3.5.)  Angle CBF plus angle CBG must therefore equal angle DBH plus angle DBG.  (C.N.2.)  Moreover, angle ABF plus angle ABI must equal angle EBH plus angle EBI.  Consequently, angles FBG, HBG, FBI, and HBI must be right because they are all equal.  (I.4.)
  4. Straight line GI is perpendicular to straight line FH.  (I. Def. 10.)

Proposition 28.4.  The opposite sides of a quadrilateral formed from the four intersections of two diameters with a circle are parallel.

  1. Angle BCG is equal to angle BEI.  (I.28.2.6.)  Likewise, angle BDG is equal to angle BAI.
  2. Angles BGC, BGD, BIA, and BIE are all equal right angles.  (I.28.3.7.)
  3. Straight lines AE and CD are parallel.  (I.27.)
  4. Angle BCF is equal to angle BEH.  (I.28.2.7)  Likewise, angle BAF is equal to angle BDH.
  5. Angles, BFA, BFC, BHD, and BHE are all equal right angles.  (I.28.3.8.)
  6. Straight lines AC and DE are parallel.

Proposition 28.5.  The opposite angles of a rhomboid are equal.

  1. Take any given rectilinear angle DAB and cut off one leg such that it is not equal to the other.  (I.3.)  Let segment AB be less than segment AD.  Let angle DAB be acute.
  2. Join points B and D by a straight line.
  3. Let segment BC be equal to segment AD.  (I.2.)  Let segment DC be equal to segment AB.  Locate point C at the point where segments equal to BC and AD intersect.  (I.7.)
  4. Triangle CDB is equal to triangle ABD because segment BD is common, segment DC is equal to segment AB, and segment BC is equal to segment AD.  (I.7.)
  5. Angle DAB is equal to angle BCD.
  6. For the same reasons, triangle ADC is equal to triangle CBA.  (I.28.5.3 and 4.)
  7. Angle ADC is equal to angle CBA.

Proposition 28.6.  If two opposite angles of a rhomboid are acute, the other two angles must be obtuse.

  1. Construct rhomboid ABCD such that angle DAB is acute and segment AB not equal to AD.
  2. Adjacent to side AD, construct equal rhomboid AEFD.  (I.28.5)
  3. Triangle ABD is equal to triangle AED because segment AD is common, segments AB and AE are equal radii of circle AB, and angle DAB is equal to angle DAE.  (I.28.6.3.) 
  4. Connect points B and E with a straight line and locate point G at the intersection of segments AD and BE.  Triangle ABG is equal to triangle AEG.  (I.4.)
  5. Angles AGB and AGE are equal and they add up to two right angles.  Therefore, they must be right angles.  (P.4.)
  6. Triangles GBD and GED are equal because segment GD is common, angles DGB and DBE are right, and segments GB and GE are equal.  (I.4.)
  1. Triangles BDC and EDF were constructed equal.  (I.28.5.4 and 6.1.)
  2. Angles ADC and ADF are equal by construction.
  3. Connect points C and F with a straight line and locate point H at the intersection of lines AD and CF.
  4. Angles HDC and HDF are equal because each is equal to two right angles less respective equal angles GDC and GDF.  (C.N.3.)
  5. Triangles HDC and DHF are equal because segment DH is common, angles HDC and HDF are equal, and segments DC and DF are equal by construction.  (I.4.)
  6. Therefore, angles DHC and DHF are right angles.  (P.4.)
  1. Angle ADC must be greater than angle DHC.  (I.16.)  But angle DHC is a right angle.  Therefore, angle ADC must be obtuse.  Angle ADF, which is equal to angle ADC, must also be obtuse.
  2. Angles CBA and FEA must also be obtuse.  (I.28.5.)

Proposition 28.7.  The opposite sides of a rhomboid are parallel.

  1. In rhomboid ABCD, the opposite angles ADC and CBA are equal.  (I.28.5.)  The opposite sides AB and DC, and AD and BC, are equal in length.  (I.28.5.3.)
  2. Triangle ABC is equal to triangle CDA.  (I.28.5.6.)
  3. Therefore, angle BCA is equal to angle DAC. 
  4. Lines AD and BC are parallel.  (I.27.)
  5. Likewise, triangles ABD and CDB are equal.  (I.28.5.4.)
  6. Angles BDC and DBA are equal.
  7. Lines AB and CD are therefore parallel.

Proposition 28.8.  The straight lines connecting the opposite points of a rhomboid are not equal.

  1. In rhomboid ABCD, let angles DAB and DCB be acute, and angles ADC and CBA be obtuse.
  2. The obtuse angles are greater than the acute angles.  (I. Def. 11 and 12.)
  3. Therefore, the sides subtending the obtuse angles must be greater.  (I.19.) 
  4. Triangles AID and CIB are not isosceles.  Likewise, triangles AIB and CID are not isosceles. 
  5. Therefore, segments AC and BD are not equal.

Proposition 28.9.  A quadrilateral formed from the four intersections of two diameters with a circle is an oblong (rectangle).

  1. Construct quadrilateral ACDE.  (I.28.1)
  2. Construct triangle AJC equal and opposite to triangle ABC.   (I.7.)
  3. Construct quadrilateral ACKL equal to quadrilateral ACDE from triangle AJC.  (I.28.1.) Segments ABCJ form a rhombus. (I. Def. 22.)
  4. I say segments EA and AL form a straight line. Triangles BDE, BCA, JCA, and JKL were all constructed equal. Point H bisects segment DE, point F bisects segment CA, and point N bisects segment KL. Line HN is straight. (I.28.3 and 28.9.3.) Lines DK, HN, and EL are all equidistant at multiple points.
  1. Suppose segments EA and AL’ form an angle other than a right angle. Let angle CAL’ be obtuse. Then straight line AL’ meets straight line CJ at point L”. If figure ACKL’ is a quadrilateral formed from the four intersections of two diameters with circle JA, then segment CL” should be equal to segment AK. (I.28.1.) But that is impossible because segment CL” is longer than segment CL (I.24), which is equal to segment AK. Also:
    • triangle AJL” is not isosceles; (I.28.2.)
    • segments JM and JO’ are not equal; (I.28.3.5)
    • point N’, which bisects points K and L”, is not the same as point N. (See I.28.3.)
  2. If angle CAL is obtuse, angle ACK must be acute, and vice versa. (I.28.6.) But triangle CAL is equal to triangle ACK–segments AK and CL are equal (I.28.1.4), segments AL and CK are equal (I.28.2), and segment AC is common. (I.7.) Thus, angles CAL and ACK are equal. Angles CAE and ACD are equal to angles CAL and ACK. (I.28.9.3.) Segments EA and AL therefore form straight lines. (I.14.) Segments DC and CK also form a straight line for the same reasons.
  1. Construct triangle APE equal and opposite to triangle ABE. (I.7.)
  2. Construct quadrilateral ASRE equal to quadrilateral ACDE from triangle APE.
  3. In the same manner, construct quadrilateral ASTL.
  1. Segments CE, CL, SE, and SL are all equal diameters of equal circles BA, JA, PA, and QA. Segments AE and AL are equal by construction. Likewise, segments AC and AS are equal by construction. Therefore, triangles ACE, ACL, ASE, and ASL are all equal. Segments EC, CL, LS, and SE form a rhombus. (I. Def. 22 and I.28.9.3.)
  2. Angles CAE, CAL, SAE, and SAL are all equal, and they add up to four right angles. Therefore, each must be a right angle.
  1. Triangle DEC is equal to triangle ACE. (I.7.) This is so because segment DE is equal to segment AC. (I.28.2.) Likewise, segment CD is equal to segment AE. Segment EC is common.
  2. Angle CAE is a right angle. Therefore, angle CDE must also be a right angle. For the same reasons, angles CAL, CKL, SAE, SRE, SAL, and STL are also right angles.
  3. Triangles ACD and AED are also equal to triangle ACE. (I.28.9.12.) Therefore, angles ACD and AED are also right angles. For the same reasons, angles ACK, ALK, ASR, AER, AST, and ALT are also right angles.
  4. Figures ACDE, ACKL, ASRE, and ASTL are all oblongs (rectangles).  (I. Def. 22.)
  1. Segment EL is straight, segment CS is straight, and segments EL and CS are perpendicular.  (I.28.9.13.)
  2. Angles EAD and LAT are equal.  (I.15.)
  3. Segment DT forms a straight line because angle CAL is right and angles CAD plus LAT are equal to a right angle (CAE).  (C.N.2.)  Likewise, segment RK is straight.
  4. Segments DT and KR are equal diameters of circle AD.
  1. Triangles DTK, KRD, DKR, and KDT are all equal because segments CD, CK, SR, and ST are all equal, segments DE, ER, KL, and LT are all equal, and radii AD, AK, AR, and AT are equal.  (C.N.2.)
  2. Angles DKT, KDR, RTK, and TRD are all right angles.  (I.28.9.13.) Lines DK, RT, DR, and KT are all straight. (I.28.9.6.)
  3. Figure DKTR is also an oblong (rectangle).

Proposition 28.10.  The three angles comprising a triangle add up to two right angles.

  1. Quadrilateral ACDE is an oblong formed from the four intersections of diameters AD and CE with circle BA.
  2. Angle ACD is a right angle.  (I.28.9.)
  3. Angle BCG is equal to angle BDG because triangle CBD is isosceles.  (I.28.2.) 
  4. Angle BCF is equal to angle BAF because triangle ABC is isosceles. 
  5. Triangle ACD is comprised of right angle ACD plus angles DAC and CDA, which together add up to a right angle. 

Proposition 28.11.  If the three angles forming two right triangles are the same, and the length of each respective side is proportional, the triangles are similar, and the sides form parallel lines.

  1. Quadrilateral ACDE is an oblong formed from the four intersections of diameters AD and CE with circle BA.
  2. Angle BDG is contained in triangles BDG, ADC, and TDK.  (I.28.9.14.)
  3. Angles CAD and KTD are equal because triangle LQT is equal to triangle ABC.  (I.28.9.2, 8, and 9.) 
  4. Angles ACD and TKD are equal because both are right angles.  (I.28.9.13 and 14.)
  5. Triangles ADC and TDK are similar, with each corresponding angle being equal and each corresponding leg of triangle TDK being double the length of that of triangle ADC.
  6. Segments AC and TK are parallel.  (I.28.)
  7. Triangle BDG can be shown to be a similar to triangle ADC with each corresponding leg being half that of triangle ADC.  

Proposition 28.12.  Parallel lines with any increment can be constructed by varying the angle of two diameters of a circle and by varying the length of the radii of the circle.

  1. Construct oblong ACDE from the four intersections of diameters AD and CE with circle BA.
  2. Lines AE and CD are parallel.  Likewise, lines AC and ED are parallel.  (I.28.4.)
  3. Construct oblong A’C’D’E’ from the four intersections of diameters A’D’ and C’E’ with circle BA’.  (I.28.2.) 
  4. Lines A’E’ and C’D’ are parallel.  Likewise, lines A’C’ and E’D’ are parallel. 

Proposition 28.13. (Fifth Postulate)  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.

  1. Let lines AB and CD be the two straight lines, and line BC be the one straight line that falls across lines AB and CD.  Angles ABC and DCB total less than two right angles.  
  1. Select point E at random along line CD.  Construct a straight line from point E that is perpendicular to line AB and that crosses line AB and point F.  (I.11.)  Produce line EF further to point G such that segment FG equals segment FE.
  1. Produce line GB to point H where it intersects circle BE.  Produce line EB to point I where it intersects circle BE. 
  1. Connect points E and H using a straight line.  Line EH is parallel to line AB.  (I.28.12.)
  2. Locate point J at the intersection of lines EH and BC.
  3. Points C, E, and J form a triangle.
  4. Because line EH is parallel to line AB, angle CJE is equal to angle CBA.
  5. Therefore, lines CE and BA will meet at some point K such that angle CKB will be equal to angle CEJ.  (I.28.12.)

Q.E.D.  Euclid’s Fifth Postulate is proven!

What is the takeaway? Circles create parallel lines. Straight lines are necessary for circles. Circles and parallel lines can both be created if Euclid’s stipulation that straight lines be used in Book I is followed. The Fifth Postulate follows from the use of straight lines and need not be assumed. Importantly, if the notion that failure to prove Euclid’s Fifth Postulate leads to a non-Euclidean reality is followed, neither circles nor parallel lines would be possible. Since circles and parallel lines are both intuitive and ubiquitous, we should not assume reality is non-Euclidean.

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