Sindbad~EG File Manager

NEAE
EEEAEC
EEEAEB
EEEAED
NECB
NEAB
NCABC
NCEAB  
NCEAC   
BEAEH
BEBHC 
ANBEAEF+EEEFAE   lemma:extension
BEAEF
EEEFAE
EEAEEF    lemma:congruencesymmetric

NCBAE   lemma:NCorder
NCAEB   lemma:NCorder
TRBAE   defn:triangle
TREAB   defn:triangle
ISEAB   defn:isosceles
EAEABEBA   proposition:05
%Therefore the sum of the angles EAB and EBA is double the angle EAB.  I.32
ASEBABAEBEF  proposition:32  
EAEBAEAB   lemma:equalanglessymmetric
EAEABBAE  lemma:ABCequalsCBA
EAEBABAE   lemma:equalanglestransitive
DABEFEBA     lemma:sumofequalsisdouble 
            %Therefore the angle BEF, is also double the angle EAB.

NCCAE   lemma:NCorder
NCAEC  lemma:NCorder
TRCAE   defn:triangle
TREAC   defn:triangle
ISEAC   defn:isosceles
EAEACECA  proposition:05
ASECACAECEF  proposition:32
EAECAEAC  lemma:equalanglessymmetric
EAEACCAE   lemma:ABCequalsCBA
EAECACAE  lemma:equalanglestransitive
DACEFECA    lemma:sumofequalsisdouble
%For the same reason the angle FEC is also double the angle EAC.
% Now we have to use point H that Euclid didn't mention

EQBB       cn:equalityreflextive
RAEBB      lemma:ray4
EQCC       cn:equalityreflexive
RAECC      lemma:ray4
RAEHF      lemma:ray4
RAEFH      lemma:ray4
EABEFBEH   lemma:equalangleshelper
EAFECHEC   lemma:equalangleshelper
EABEHBEF   lemma:equalangelssymmetric
EAHECFEC   lemma:equalanglessymmetric
ASBEFBEBBEC   defn:anglesum

RAABB    lemma:ray4
RAACC    lemma:ray4
RAAHF    lemma:ray4
RAAFH    lemma:ray4
EABAHBAF  lemma:equalangleshelper
EACAHCAF  lemma:equalangleshelper
EABAFBAH  lemma:equalanglessymmetric
EACAFCAH  lemma:equalanglessymmetric
ASBAFFACBAC  defn:anglesum







case 1: ASBAFFACBAC
  % and then we can show also
 ASBEFFECBEC
  % and thus since AEBEFFEC also
 ASBEFBEFBEC 
 ASBAFBACBAC
  % and hence
 DABECBEF  lemma:doublesum


Again let another straight line be inflected, and let there be another angle BDC. Join DE and produced it to G.
Similarly then we can prove that the angle GEC is double the angle EDC, of which the angle GEB is double the angle EDB. Therefore the remaining angle BEC is double the angle BDC.	
DABECBDC