Răspuns:
Explicație pas cu pas:
MN║AC, ⇒ΔMBE~ΔABO. Fie ME=x=OE, atunci MN=2x. Din ⇒ΔMBE~ΔABO, ⇒[tex]\dfrac{ME}{AO}=\dfrac{BE}{BO},~=>~\dfrac{x}{5\sqrt{3} }=\dfrac{5-x}{5}~|*5,~=>~\dfrac{x}{\sqrt{3} }=\dfrac{5-x}{1}~=>~x*1=\sqrt{3}*(5-x),~=>~x*1=5\sqrt{3}-\sqrt{3}*x,~=>~\sqrt{3}*x+x*1=5\sqrt{3},~=>~x*(\sqrt{3}+1)=5\sqrt{3},~=>~x=\dfrac{5\sqrt{3}}{\sqrt{3}+1}=\dfrac{5\sqrt{3}(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\dfrac{5\sqrt{3}(\sqrt{3}-1)}{(\sqrt{3})^2-1^2}=\dfrac{5\sqrt{3}(\sqrt{3}-1)}{3-1}=\dfrac{5\sqrt{3}(\sqrt{3}-1)}{2}=[/tex]
Deci, 2x=5√3(√3-1)=5((√3)²-√3)=5(3-√3)cm=MN.
