ΔABC - isoscel, AC = BC
ΔABD -isoscel, AB = AD = √5 cm
ΔADC -isoscel, AD= CD = √5 cm
[tex]\Delta ABC\sim\ \Delta BDA \Rightarrow\dfrac{AB}{BD}=\dfrac{BC}{AD}\\ \\ \dfrac{\sqrt5}{BD}=\dfrac{BC}{\sqrt5} \Rightarrow BD\cdot BC=\sqrt5\cdot\sqrt5\\ \\ (BC-\sqrt5)\cdot BC=5 \Rightarrow BC^2-\sqrt5BC-6=0[/tex]
[tex]\it \Rightarrow BC=\dfrac{\sqrt5+5}{2}\ cm \Rightarrow AC=\dfrac{\sqrt5+5}{2}\ cm\\ \\ \\ \mathcal{P} =AB+BC+CA = \sqrt5+\dfrac{\sqrt5+5}{2}+\dfrac{\sqrt5+5}{2}=2\sqrt5+5\ cm[/tex]
