Prob 1)
Δdnm                    Δcnm

dn              =          nc   (as n is d mid point of dc)
angle mnd =          angle mnc (as mn perpendicular to dc)
mn             =          mn (common side)

by s.a.s. axiom Δdnm congruent to Δcnm

since Δdnm congruent to Δcnm dm=cm

2)  Δadb                  Δadc
    angle adb    =      angle adc (ad perpendicular to bc)
    angle dab    =       angle dac (given)
    ad                =        ad (common side)
by a.s.a. axiom  Δadb congruent to Δadc
therefore angle abd = angle acd(corresponding angles of Δadb n Δadc)
therefore abc is a isosceles triangle

3) Δabd                     Δacd
 ab              =             ac
 ad              =             ad
angle adb   =             angle adc

by s.a.s. axiom  Δabd congruent to Δacd

angle b = angle c ( corresponding angles of Δabd n Δacd)
ad bisects bc since Δabd congruent to Δacd