a) AD -bisectoare ⇒ Teorema bisectoare- AB/AC=BD/DC
BD/DC= 25/20=5/4 = k
BD=5k si CD=4k , BD+CD=BC
5k+4k=18
9k=18
k=2
BD=5*2=10 cm si CD= 4*2=8cm
b) AE=AB-BE , BE=BD=10cm
AE=25-10 = 15cm
AF=AC- CF , CF=DC = 8cm
AF=20-8=12cm
AE/AB=AF/AC⇔ 15/25= 12/20 = 3/5 (A)⇒ RTTH⇒ AF||BC
⇒ΔABC~ΔAEF⇒ AE/AB=AF/AC=EF/BC⇒ 3/5 = EF/18
EF=18*3:5= 10,8cm
P AEF =AE+EF+AF= 15+12+10,8=37,8cm