a)
[tex]\it AD=AB=24\ cm \Rightarrow \Delta ABD-isoscel \Rightarrow \widehat{BDA}=\widehat{ABD}=75^o\\ \\ \widehat A=180^o-2\cdot75^o=30^o\\ \\ \mathcal{A}_{ABCD}=AB\cdot AD\cdot sinA=24\cdot24\cdot sin30^o=24\cdot24\cdot\dfrac{1}{2}=288\ cm^2[/tex]
b)
[tex]\it ABCD-romb \Rightarrow \widehat{ABC}=150^o\ (suplementul \ unghiului\ \widehat A=\ 30^o)\ \ \ \ \ (1)\\ \\ M\ -\ mijlocul\ lui\ [AB]\ \ \ \ \ (2)\\ \\ E\ -\ simetricul\ lui \ D\ fa\c{\it t}\breve a\ de\ M \Rightarrow \ M\ -\ mijlocul\ lui\ [DE]\ \ \ \ \ (3) \\ \\ (2),\ (3) \Rightarrow AEBD-paralelogram \Rightarrow AE||BD,\ AB\ -\ secant\breve a \Rightarrow \\ \\ \Rightarrow \widehat{EBA}=\widehat{DAB}=30^o\ (alterne\ interne)\ \ \ \ \ (4)\\ \\ (1),\ (4) \Rightarrow \widehat{EBC}=150^o+30^o=180^o[/tex]
Deci, punctele E, B, C sunt coliniare
