Rysunek pomocniczy:

Z treści zadania wiemy, że:

$P_{ABLK}=P_{KLC}$  

 

Zauważmy, że:    

$P_{ABC}=P_{KLC}+P_{ABLK}$  

$P_{ABC}=2\cdot P_{KLC}$  

$\frac{P_{ABC}}{2}=P_{KLC}$  

$\frac{1}{2}{P}_{ABC}=P_{KLC}$  

$\frac{1}{2}=\frac{P_{KLC}}{P_{ABC}}$  

 

Obliczmy skalę podobieństwa trójkątów KLC i ABC.

$\frac{P_{KLC}}{P_{ABC}}=k^2$  

$\frac{1}{2}=k^2$  

$\frac{1}{\sqrt{2}}=k$  

$\frac{\sqrt{2}}{2}=k$  

 

Zatem otrzymujemy:

$\frac{\left|CK\right|}{\left|CA\right|}=\frac{\sqrt{2}}{2}$  

$2\cdot \left|CK\right|=\sqrt{2}\cdot \left|CA\right|$  

$2\cdot \left|CK\right|=\sqrt{2}\cdot \left(\left|CK\right|+\left|KA\right|\right)$  

$2\cdot \left|CK\right|=\sqrt{2}\left|CK\right|+\sqrt{2}\left|KA\right|\left|-\sqrt{2}\right|CK\mid$  

$2\left|CK\right|-\sqrt{2}\left|CK\right|=\sqrt{2}\left|KA\right|$  

$\left|CK\right|\left(2-\sqrt{2}\right)=\sqrt{2}\left|KA\right|$  

$\frac{\left|CK\right|}{\left|KA\right|}=\frac{\sqrt{2}}{2-\sqrt{2}}$  

$\frac{\left|CK\right|}{\left|KA\right|}=\frac{\sqrt{2}\left(2+\sqrt{2}\right)}{2^2-{\sqrt{2}}^2}$  

$\frac{\left|CK\right|}{\left|KA\right|}=\frac{2\sqrt{2}+2}{4-2}$  

$\frac{\left|CK\right|}{\left|KA\right|}=\frac{2\sqrt{2}+2}{2}$  

$\frac{\left|CK\right|}{\left|KA\right|}=\sqrt{2}+1$