In $△\mathrm{ADC}$ and $△\mathrm{BAC}$,
$\angle ADC=\angle BAC$ (Already given)
$\angle \mathrm{ACD}=\angle \mathrm{BCA}$ (Common angles)
$\therefore △\mathrm{ADC}\sim △\mathrm{BAC}$ (AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
$ \displaystyle \dfrac{{\text{CA}}}{{\text{CB}}}=\dfrac{{\text{CD}}}{{\text{CA}}}$
​∴ $ \displaystyle {\Rightarrow \text{C}{{\text{A}}^{2}}=\text{CB}\text{.CD}\text{.}}$ 
Hence, proved.