[tex]\it Not\breve am\ BC=x \Rightarrow AB=2x\\ \\ Cu\ teorema\ lui\ Pitagora\ \hat\imath n\ \Delta ABC \Rightarrow AC=x\sqrt3\\ \\ \Delta CAB\sim\Delta DCA \Rightarrow \dfrac{AB}{AC}=\dfrac{AC}{CD} \Rightarrow CD=\dfrac{AC^2}{AB}=\dfrac{3x^2}{2x}=1,5x\\ \\ \\ PQ=3 \Rightarrow \dfrac{AB-CD}{2}=3 \Rightarrow AB-CD=6 \Rightarrow 2x-1,5x=6 \Rightarrow \\ \\ \\ \Rightarrow 0,5x=6|_{\cdot2} \Rightarrow x=12 \Rightarrow AB=24cm,\ CD=18\ cm\\ \\ \\ MN=\dfrac{AB+CD}{2}=\dfrac{24+18}{2}=21\ cm[/tex]
