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Нека [tex]\angle ACD = \alpha ,\angle ADC = \beta ,\angle CAD = \gamma[/tex]

За $\triangle ACD$ прилагам Синусова теорема:
[tex]\frac{AD}{sin\alpha} = 2R \Leftrightarrow AD = sin\alpha.2R \Leftrightarrow AD = \frac{\sqrt{15}}{8}.\frac{16\sqrt{15}}{15}\Rightarrow AD = 2[/tex]
$\frac{AC}{sin\beta} = 2R \Leftrightarrow \frac{4}{sin\beta} = \frac{16\sqrt{15}}{15} \Leftrightarrow sin\beta = \frac{15}{4\sqrt{15}} \Rightarrow sin\beta = \frac{\sqrt{15}}{4}$
$\frac{DC}{sin\gamma} = 2R \Leftrightarrow \frac{DC}{sin[180^\circ- (\alpha + \beta)]} = \frac{16\sqrt{15}}{15} \Leftrightarrow DC = sin(\alpha + \beta).\frac{16\sqrt{15}}{15}$
$sin(\alpha + \beta) = sin\alpha.cos\beta + sin\beta.cos\alpha$
$sin\alpha = \frac{\sqrt{15}}{8} , cos\alpha = \frac{7}{8}$
$sin\beta = \frac{\sqrt{15}}{4} ,cos\beta = \frac{1}{4}$
Тогава $sin\gamma = sin(\alpha + \beta) = \frac{\sqrt{15}}{4}$
$DC= \frac{\sqrt{15}}{4}.\frac{16\sqrt{15}}{15} \Rightarrow DC = 4$
$$AB = DC = 4$$
$S_{ABCD } = AD.CD.sin\beta \Leftrightarrow S_{ABCD } = 2.4.\frac{\sqrt{15}}{4}$
$$S_{ABCD } = 2\sqrt{15}$$