a) We have : DM = HD

                     BD = CD

               \(\widehat{BDM}=\widehat{CDH}\)  ( 2 Vertical angles )

=> \(\Delta BDM=\Delta CDH\)   (1) 

b) From (1) , we have :

\(\widehat{HCB}=\widehat{MBC}\)

But \(\Delta ABC\) is the isosceles triangle

So that \(\widehat{HCB}=\widehat{ABC}\)

=> \(\widehat{MBC}=\widehat{ABC}\)

=> BC is the bisector of \(\widehat{ABM}\)

c) I don't understand the thread