ΔADB dr. in D
ung. BAD=30⇒ DB=AB/DB
DB=[tex] \frac{8 \sqrt{3} }{2} [/tex]
DB=[tex]4 \sqrt{3} [/tex]
⇒T.P. [tex]AD^{2} +DB^{2} =AB^{2} [/tex]
[tex]AD ^{2} +(4 \sqrt{3})^{2} =(8 \sqrt{3} )^{2} [/tex]
[tex]AD^{2} =192-48[/tex][tex] AD^{2} =144[/tex] AD=12
ΔADC dr. in D
ung. ACD=30
⇒AD=AC/2
[tex]12= \frac{AC}{2} [/tex]
AC=24
IN TRIUNGHIUL ABC dr. in A⇒ T.P.
[tex]AB^{2} +AC^{2} =BC^{2}
[/tex]
[tex]24^{2} +(8 \sqrt{3})^{2} =BC^{2} [/tex]
576+192=[tex]BC^{2} [/tex]
[tex]BC^{2} =768[/tex]
BC=[tex]16 \sqrt{3} [/tex]
