[tex]\it \left.\begin{aligned}\it BD\ -\ bisectoare\ pentru\ \hat B \Rightarrow \hat B_1=\hat B_2\\ \\ \it AB||CD,\ \ \hat B_2=\hat D_1\ (alterne\ \ interne)\end{aligned}\right\}\ \Rightarrow\ \hat B_1=\hat D_1 \Rightarrow\Delta BCD-isoscel\Rightarrow\\ \\ \\ \Rightarrow BC=CD=100m[/tex]
[tex]\it Ducem\ DF||BC \Rightarrow FBCD-paralelogram\ (are\ laturile\ opuse\ paralele).\\ \\ Dar, BC=CD \Rightarrow FBCD\ -\ romb \Rightarrow DF=FB=BC=CD=100m[/tex]
[tex]\it AF=AB-FB=160-100=60m\\ \\ \Delta DAF-dreptunghic,\ \hat A=90^o,\ \stackrel{T.Pitagora}{\Longrightarrow}\ \ AD^2=DF^2-AF^2=100^2-60^2=\\ \\ =10000-3600=6400=80^2 \Rightarrow AD=80m=h\ \ (pentru\ \ trapez)[/tex]
[tex]\it \mathcal{A}=\dfrac{\mathcal{B}+b}{2}\cdot h=\dfrac{AB+CD}{2}\cdot AD=\dfrac{160+100}{2}\cdot80=130\cdot80=10400\ m^2[/tex]
