A,B,C,D coliniare
5AB=9AC-4AD; BD=18
a.BD=AD-AB sau AD=AB+BD
AB+BC=AC
5AB=9(AB+BC)-4(AB+BD)
5AB=9AB+9BC-4AB-4BD
5AB+4AB+4BD=9AB+9BC
9AB+4BD=9AB+9BC
|-9AB
4BD=9BC
BC=4BD:9
BC=4×18:9
BC=4×2
BC=8 cm
BC+CD=BD
CD=BD-BC
CD=18-8
CD=10 cm
b. P - mijlocul lui AD, P∈(BC)
BD=18
AD are valoarea maxima posibil daca AP=PB=17,5
=>AD=17,5·2; AD=35, e nr natural
AD=35 cm
