[tex]BD^2=AB^2+AD^2-2AD\cdot AB\cos120=64+16-64(-\frac{1}{2})=\\
80+32=112 \Rightarrow BD=\sqrt{112}=4\sqrt7\\
AC^2=BA^2+BC^2-2BA\cdot BD\cos 60=64+16-64\frac{1}{2}=48\\
AC=\sqrt{48}=4\sqrt3\\
\text{Problema 2}
\\ \text{Aflam lungimea diagonalei $AC$ cu teorema lui Pitagora si obtinem}\\
AC=25\\
\text{$DE$ este inaltime in triunghiul dreptunghic $DAC$}\\
DE=\frac{DA\cdot DC}{AC}=\frac{15\cdot20}{25}=12\\
AE=\frac{AD^2}{AC}=\frac{15^2}{25}=9
\\
\Delta AEF\sim\Delta CDA\Rightarrow \\
[/tex]
[tex]\frac{AE}{DC}=\frac{EF}{AD}\Rightarrow EF=\frac{15\cdot 9}{20}=\frac{27}{4}\\
DF=DE+EF=12+\frac{27}{4}=\frac{75}{4}[/tex]
[tex]BG=GC-BC\\
\text{Pe $GC$ il scoti din asemanrea }\Delta ADC\sim\Delta DCG [/tex]
