ad²+dc²=ac²
24²+(8√3)²=ac²
576+192=ac²
ac²=768⇒ac=√768⇒ac=16√3
de perpendicular ac
aplic teorema catetei in Δdac
dc²=ac×ec
(8√3)²=16√3×ec
192=16√3×ec⇒ec=192/16√3⇒ec=13/√3⇒ec=12√3/3⇒ec=4√3
aplic teorema inaltimii in Δdac
ed²=ec×ae
ae=ac-ec⇒ae=16√3-4√3⇒ae=12√3 din ultimele doua ⇒
ed²=4√3×12√3
ed²=4×12×3
ed²=144⇒ed=√144⇒ed=12
in Δfcd
ec=4√3 si ecperpendicular df
de=12
teorema inaltimii si din toate trei ⇒
ec²=de×ef
(4√3)²=12×ef
48=12×ef⇒ef=48/12⇒ef=4
df=de+ef
df=12+4
df=16
pitagora: dc²+fc²=df²
fc²=df²-dc²
fc²=16²-(8√3)²
fc²=256-192
fc²=64⇒fc=√64⇒fc=8
arie cef=ce×ef/2=4√3×4/2=8√3
arie dec=de×ec/2=12×4√3/2=24√3
arie cef/arie dec=8√3/24√3=8/24=1/3
