Trójkąt ABC jest równoramienny, więc:

|AC|=|BC|=x 

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Ze wzoru na pole trójkąta:

$P_{ACD}=\frac{1}{2}x\cdot \left|CD\right|\cdot \sin \alpha$  

$P_{BCD}=\frac{1}{2}x\cdot \left|CD\right|\cdot \sin \beta$  

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Mamy więc:

$\frac{P_{ACD}}{P_{BCD}}=\frac{\frac{1}{2}x\cdot \left|CD\right|\cdot \sin \alpha }{\frac{1}{2}x\cdot \left|CD\right|\cdot \sin \beta }=\frac{\sin \alpha }{\sin \beta }$  

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Oznaczmy:

$\left|\sphericalangle CDB\right|=\gamma$  

Wówczas:

$\left|\sphericalangle ADC\right|={180}^{\circ }-\gamma$  

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Ze wzoru na pole trójkąta:

$P_{ADC}=\frac{1}{2}\left|AD\right|\cdot \left|CD\right|\cdot \sin \left({180}^{\circ }-\gamma \right)=\frac{1}{2}\left|AD\right|\cdot \left|DC\right|\cdot \sin \gamma$  

$P_{BCD}=\frac{1}{2}\left|DB\right|\cdot \left|CD\right|\cdot \sin \gamma =\frac{1}{2}\left|DB\right|\cdot \left|CD\right|\cdot \sin \gamma$  

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Zatem:

$\frac{P_{ACD}}{P_{BCD}}=\frac{\frac{1}{2}\left|AD\right|\cdot \left|CD\right|\cdot \sin \gamma }{\frac{1}{2}\left|DB\right|\cdot \left|CD\right|\cdot \sin \gamma }=\frac{\left|AD\right|}{\left|DB\right|}$  

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Otrzymaliśmy:

$\frac{\sin \alpha }{\sin \beta }=\frac{P_{ACD}}{P_{BCD}}=\frac{\left|AD\right|}{\left|DB\right|}$