My drawing is not accurate ha ! ^ ^ Sorry about that !

Draw BX parallel to AC

=> \(\widehat{BXE}=\widehat{CEX}\)

Consider \(\Delta BMX\) and \(\Delta CEM\) , we have :

BM = CM

\(\widehat{BMX}=\widehat{CME}\)  => \(\Delta BMX\) = \(\Delta CEM\)

\(\widehat{BXE}=\widehat{CEX}\)

Consider \(\Delta AHD\) and \(\Delta AHE\) , we have :

\(\widehat{DAH}=\widehat{CAH}\)

AH general                     => \(\Delta AHD\) = \(\Delta AHE\)

\(\widehat{AHD}=\widehat{AHE}=90^0\)

=> \(\widehat{ADE}=\widehat{AED}\)

But \(\widehat{AED}=\widehat{BXD}\)

=> \(\widehat{ADE}=\widehat{BXD}\)

=>\(\Delta\)BDX  isosceles 

=> BD = BX

But BX = CE

=> BD = CE