Question

In the given figure, $\angle A={90}^{\circ }$ and $AD\perp BC$ If $BD=2cm$ and $CD=8cm$, find $AD$.

Answer 1

From the figure, consider $△ABC$,

So, $\angle A={90}^{\circ }$

And $AD\perp BC$

$\angle BAC={90}^{\circ }$

Then, $\angle BAD+\angle DAC={90}^{\circ }$ … [equation (i)]

Now, consider $△ADC$

$\angle ADC={90}^{\circ }$

So, $\angle DCA+\angle DAC={90}^{\circ }$ … [equation (ii)]

From equation (i) and equation (ii)

We have,

$\angle BAD+\angle DAC=\angle DCA+\angle DAC$

$\angle BAD=\angle DCA$ … [equation (iii)]

So, from $△BDA$ and $△ADC$

$\angle BDA=\angle ADC$ … [both the angles are equal to ${90}^{\circ }$]

$\angle BAD=\angle DCA$ … [from equation (iii)]

Therefore, $\angle BDA\sim \angle ADC$

$BD/AD=AD/DC=AB/AC$

Because, corresponding sides of similar triangles are proportional

$BD/AD=AD/DC$

By cross multiplication we get,

$A{D}^2=BD\times CD$

$A{D}^2=2\times 8=16$

$AD=\sqrt{16}$

$AD=4$.