a) \(\Delta ABE\) and \(\Delta ACD\) have the common angle A ; AB = AC ; AE = AD, so \(\Delta ABE=\Delta ACD\left(S-A-S\right)\)

b) From a), we have : \(\widehat{B_1}=\widehat{C_1}\). Moreover, \(\widehat{B_1}=\widehat{A_1}\)(since they're both complementary to \(\widehat{A_2}\)). So, \(\widehat{C_1}=\widehat{A_1}\). Hence, \(\Delta MAC\) isosceles at M

c) From b), we have MA = MC (1)

Since \(\widehat{C_1}+\widehat{D_1}=90^0;\widehat{A_1}+\widehat{A_2}=90^0;\widehat{C_1}=\widehat{A_1}\), \(\widehat{D_1}=\widehat{A_2}\). Hence, \(\Delta MAD\) isosceles at M. Then, MA = MD (2)

(1),(2) => MC = MD => M is the midpoint of CD

\(DI\perp BE;AK\perp BE\Rightarrow\)DI // AK

\(\Delta DCI\) has MC = MD ; MK // DI, so K is the midpoint of IC

d) The medians DK, IM of \(\Delta DCI\) intersect at G, so G is the centroid of \(\Delta DCI\). CG cuts DI at N, then ND = NI (3)

\(\Delta DNC\) has MC = MD ; MF // DN, so F is the midpoint of NC

=> MF is the midsegment of \(\Delta DNC\Rightarrow MF=\dfrac{DN}{2}\) (4)

Similarly, we have : \(FK=\dfrac{NI}{2}\left(5\right)\)

(3),(4),(5) => FM = FK

I can't draw this triangle.