construim BM perpendicular pe DC. BM=AD =15
in ΔBMC ⇒tg C=[tex] \frac{BM}{MC} [/tex] ⇒[tex] \frac{6}{10} = \frac{15}{MC} \\ MC= \frac{10*15}{6} =25[/tex]
plicam th. pitagora ⇒[tex] BC^{2} = BM^{2} + MC^{2} = 15^{2} + 25^{2} =225+625=850 \\ BC= \sqrt{850=} =5 \sqrt{34} [/tex]
