[tex]\it \dfrac{DC}{BC}=\dfrac{1}{4} \Rightarrow DC=x,\ BC=4x,\ x\in\ \mathbb{R}\\ \\ BD=BC-CD=4x-x=3x\\ \\ \Delta ABC,\ T.h. \Rightarrow AD^2=BD\cdot DC=3x\cdot x=3x^2 \Rightarrow AD=x\sqrt3\ cm\\ \\ \Delta DAC-dr,\ \widehat{D}=90^o,\ \stackrel{T.\ Pitagora}{\Longrightarrow}\ \ AD^2+CD^2=AC^2 \Rightarrow \\ \\ \Rightarrow (x\sqrt3)^2+x^2=6^2 \Rightarrow 3x^2+x^2=36 \Rightarrow 4x^2=36|_{:4} \Rightarrow \\ \\ \Rightarrow x^2=9 \Rightarrow x^2=3^2 \Rightarrow x=3[/tex]
Deci:
[tex]\it CD=3cm,\ \ BD=3\cdot3=9cm,\ \ BC=4\cdot3=12cm, AD=3\sqrt3\ cm\\ \\ \mathcal{A}=\dfrac{BC\cdot AD}{2} =\dfrac{12\cdot3\sqrt3}{2}=6\cdot3\sqrt3=18\sqrt3\ cm^2[/tex]
