In given figure, if ∠ACB = ∠CDA,

Given, AC = 8 cm, AD = 3cm and            ∠ACB = ∠CDA

From figure,                                                ∠CDA = 90°

∠ACB = ∠CDA = 90°

In right angled $\triangle A D C$, $A C^{2}=A D^{2}+C D^{2}$

$\Rightarrow \quad(8)^{2}=(3)^{2}+(C D)^{2}$

$\Rightarrow \quad 64-9=C D^{2}$

$\Rightarrow \quad C D=\sqrt{55} \mathrm{~cm}$

In $\triangle C D B$ and $\triangle A D C$, $\angle B D C=\angle A D C$ [each 90°]

$\angle D B C=\angle D C A \quad$ [each equal to $90^{\circ}-\angle A$ ]

$\therefore$ $\triangle C D B \sim \Delta A D C$

Then, $\frac{C D}{B D}=\frac{A D}{C D}$

$\Rightarrow$ $C D^{2}=A D \times B D$

$\therefore$ $B D=\frac{C D^{2}}{A D}=\frac{(\sqrt{55})^{2}}{3}=\frac{55}{3} \mathrm{~cm}$