[tex]\bf AD\cap BC=\{M\}\\ \\ \ \ \ CD -linie\ mijlocie\ \hat\imath n\ \Delta MAB \Rightarrow MA=32cm,\ MB=24cm\\ \\ (24,\ 32,\ 40)\ -\ triplet\ pitagoreic \Rightarrow \Delta MAB-\ dreptunghic,\ \widehat M=90^o\\ \\ \mathcal{P}_{MAB}=24+32+40=96\ cm\\ \\ cos(ABC)=cos(ABM)=\dfrac{MB}{AB}=\dfrac{\ \ 24^{(8}}{40}=\dfrac{3}{5}\\ \\ \\ sin(DAB)=sin(MAB)=cos(90^o-\widehat{MAB})=cos(ABM)=\dfrac{3}{5}[/tex]
