[tex]
\it AD=25\%\ din\ BC =\dfrac{\ \ 25^{(25}}{100}\cdot BC =\dfrac{1}{4} \cdot BC =\dfrac{BC}{4} \ \ \ \ \ (1)
[/tex]
[tex]\it DC - BD = 4 \ \ \ \ \ (2)
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\{BD,\ DC \} \ d.\ p.\ \{ 2,\ 3 \} \Rightarrow \{DC,\ BD \} \ d.\ p.\ \{ 3,\ 2 \} \Rightarrow
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\Rightarrow \dfrac{DC}{3} =\dfrac{BD}{2} =\dfrac{DC-BD}{3-2} \stackrel{(2)}{=} \dfrac{4}{1} =4 \ \ \ \ [/tex]
[tex]\it \dfrac{DC}{3} =4 \Rightarrow DC=3\cdot4 \Rightarrow DC=12\ cm
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\dfrac{BD}{2} =4 \Rightarrow BD=2\cdot4 \Rightarrow BD =8\ cm
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BC = BD+DC = 8+12=20\ cm \ \ \ \ \ (3)[/tex]
[tex]\it (1), (3) \Rightarrow AD = \dfrac{20}{4} \Rightarrow AD =5\ cm
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\mathcal{A} = \dfrac{BC\cdot AD}{2} = \dfrac{20\cdot 5}{2} = = 50\ cm^2[/tex]
