In figure below, $ \mathrm{ABC} $ is a triangle right angled at $ \mathrm{B} $ and $ \mathrm{BD} \perp \mathrm{AC} $. If $ \mathrm{AD}=4 \mathrm{~cm} $, and $ C D=5 \mathrm{~cm} $, find $ B D $ and $ A B $.
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Given:

\( \mathrm{ABC} \) is a triangle right angled at \( \mathrm{B} \) and \( \mathrm{BD} \perp \mathrm{AC} \).

\( \mathrm{AD}=4 \mathrm{~cm} \), and \( C D=5 \mathrm{~cm} \)

To do:

We have to find \( B D \) and \( A B \).

Solution:

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

$\angle A D B =\angle C D B=90^o$

$\angle B A D =\angle D B C=90^o-\angle C$

Therefore, by AA similarity,

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

This implies,

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

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

$D B^{2} =4 \times 5$

$D B =2 \sqrt{5} \mathrm{~cm}$

In $\triangle BDC$,

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

$B C^{2}=(2 \sqrt{5})^{2}+(5)^{2}$

$=20+25$

$=45$

$B C=\sqrt{45}$

$=3 \sqrt{5}$

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

This implies,

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

$\frac{2 \sqrt{5}}{5}=\frac{B A}{3 \sqrt{5}}$

$B A=\frac{2 \sqrt{5} \times 3 \sqrt{5}}$

Hence $BD=2 \sqrt{5} \mathrm{~cm}$ and $BA=6 \mathrm{~cm}$.