In figure below, if $ \angle \mathrm{ACB}=\angle \mathrm{CDA}, \mathrm{AC}=8 \mathrm{~cm} $ and $ \mathrm{AD}=3 \mathrm{~cm} $, find $ \mathrm{BD} $.
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Given:

\( \angle \mathrm{ACB}=\angle \mathrm{CDA}, \mathrm{AC}=8 \mathrm{~cm} \) and \( \mathrm{AD}=3 \mathrm{~cm} \)

To do:

We have to find \( \mathrm{BD} \).

Solution:

$\angle \mathrm{CDA}=90^{\circ}$

$\angle \mathrm{ACB}=90^{\circ}$

$\angle \mathrm{CDA}=90^{\circ}$

In right angled triangle $ADC$,

$A C^{2}=A D^{2}+C D^{2}$

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

$CD^2=64-9$

$C D=\sqrt{55}$

In $\triangle C D B$ and $\triangle A D C$,

$\angle B D C=\angle A D C$

$\angle D B C=\angle D C A$

Therefore, by AA similarity,

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

This implies,

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

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

$B D=\frac{C D^{2}}{A D}$

$=\frac{(\sqrt{55})^{2}}{3}$

$=\frac{55}{3} \mathrm{~cm}$

Hence, $BD=\frac{55}{3} \mathrm{~cm}$