Here is my solution : a) BD = BE ; CD = CE so BC is the midperpendicular of DE. We have angle ZDE = 1/2.angle ZCE = 1/2.(180* - angle ACB - angle BCE) = 1/2.(180* - angle ABC - angle BCD) = 1/2.(180*-90*) = 45*. b) We have angle MBE = 2.angle MDE = 2.angle ZDE = 90*. So angle BEM = angle BME = (180* - angle MBD)/2 = 45*. Hence angle CEM = angle BEC - angle BEM = angle BDC - angle BEM = 90* - 45* = 45*(1). On the other hand, we have angle CEK = angle CKE = angle ZDE = 45*(2). According (1) and (2) we have E,M,K lie on a line. c) According b) we have BME = 45*; CME = 45* so BM//CE.

a) we have $\widehat{MDC}$ + $\widehat{BDE}$ = $\frac{\widehat{DBM}+180-\widehat{DBE}}{2}$ = $\frac{180-\widehat{MBE}}{2}$=$\widehat{MDE}$=$\widehat{ZDE}$ thus $90=2*\widehat{ZDE} => \widehat{ZDE}=45$. b) we have $\widehat{DEM}$ = $\widehat{CDM}$ = $\widehat{DZC}$ = $\widehat{DEK}$ therefore $E$,$M$ and $K$ lie on line c) we have $\widehat{BME}$ = 45 = $\widehat{KEC}$ = $\widehat{MEC}$ .So $BM//EC$

Nice question and beautiful solutions.