a, Xét ΔABC: góc BAC = 90 độ (gt)
=> BC^2 = AC^2 + AB^2 (Pytago)
=> BC^2 = 12^2 + 9^2
=> BC^2 = 225
=> BC = 15 (cm)
Xét ΔABC: AD phân giác góc BAC (gt)
=> BD/AB = DC/AC 
=> BD/AB = DC/AC = (BD + DC)/(AB + AC) = 15/21 = 5/7
=> BD = 5/7.AB = 5/7.9 = 45/7 (cm)
DC = 5/7.AC = 5/7.12 = 60/7 (cm)
Có AB vuông góc AC (gt)
DE vuông góc AC (gt)
=> AB // DE => DE/AB = DC/BC => DE/9 = (60/7)/15 = 4/7
=> DE = 4/7 . 9 = 36/7 (cm)
b, Ta có: SABC = 1/2.AB.AC = 1/2.9.12 = 54 (cm^2)
SADC = 1/2.DE.AC = 1/2.12.36/7 = 216/7 (cm^2)
=> SABD = SABC - SADC = 54 - 216/7 = 162/7 (cm^2)