[tex]\it a)\ AC = diagonala\ p\breve{a}tratului \Rightarrow AC=AB\sqrt2 \Rightarrow AC = 16\sqrt2\ cm\ \ \ (*)\\ \\ AN=3NC \Rightarrow \dfrac{AN}{NC}=\dfrac{3}{1}\ \stackrel{derivare}{\Longrightarrow}\ \dfrac{AN+NC}{NC}=\dfrac{3+1}{1} \Rightarrow\\ \\ \\ \Rightarrow \dfrac{AC}{NC}=\dfrac{4}{1} \stackrel{(*)}{\Longrightarrow}\ \dfrac{16\sqrt2}{NC}=\dfrac{4}{1} \Rightarrow NC=\dfrac{1\cdot16\sqrt2}{4} \Rightarrow NC=4\sqrt2\ cm[/tex]
[tex]\it b)\ \ AC=16\sqrt2cm,\ NC=4\sqrt2cm \Rightarrow NC=\dfrac{AC}{4}.\\ \\ Fie\ O,\ mijlocul\ diagonalei\ AC \Rightarrow OC = \dfrac{AC}{2}\Rightarrow ON=NC=\dfrac{AC}{4}.\\ \\ Ducem\ NF\perp MB, F\in MB \Rightarrow NF-linie\ mijlocie\ \^{i}n\ trapezul\ BCOM.\\[/tex]
Deoarece BCOM-trapez dreptunghic ⇒ NF- mediatoarea
segmentului [MB] ⇒ ΔNMB -isoscel, NM=NB.
