a)
DC=BC-BD=24-6=18cm
AD=[tex]\sqrt{BD*DC}=\sqrt{6*18}=\sqrt{6*6*3}=6\sqrt{3}[/tex]cm
[tex]A_{ADC} =\frac{AD*DC}{2} =\frac{6\sqrt{3} *18}{2} =6\sqrt{3} *9=54\sqrt{3}[/tex]cm
DE este mediana in ΔADC => [tex]A_{ADE} =A_{ACE} =\frac{A_{ADC}}{2} =\frac{54\sqrt{3} }{2} =27\sqrt{3}[/tex]
[tex]A_{ADB} =\frac{AD*DB}{2} =\frac{6\sqrt{3} *6}{2} =6\sqrt{3} *3=18\sqrt{3}[/tex]
[tex]A_{ABDE} =A_{ABD} +A_{ADE} =27\sqrt{3}+18 \sqrt{3} =45\sqrt{3}[/tex]
b)
Aplicam teorema lui pitagora in triunghiul ADC
AC²=AD²+DC²=(6√3)²+18²=108+324=432
AC=12√3cm
AE=EC=AC/2=6√3cm
In triunghiul ADE AD=DE=AE=6√3cm=>ΔADE este echilsteral
=>∡DAE=∡AED=60°
∡DAE=∡AED=60°,∡BAC=∡ADC=90°,AD=AE =>(U.L.U)ΔADC≡ΔEAF
=>AF=DC
