Question

In the given figure,$\angle$ ABC = 90 and B D $\perp$ AC. If BD = 8 cm and AD = 4 cm, find CD.

Answer 1

Given:

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

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

Let $\angle BCD=x^{\circ }$

Using triangle sum property in $△BDC$, $\angle DBC=90-x^{\circ }$

Also $\angle ABD=x^{\circ }$

Using triangle sum property in $△ADB$, $\angle BAD=90-x^{\circ }$

Now considering $△BDC$ and $△ADB$

$\angle BDC=\angle BDA$ $[\because \angle BDC=\angle BDA={90}^{\circ }]$

$\angle DBC=\angle DAB$ $[\because \angle DBC=\angle DAB=(90-x{)}^{\circ }]$

So by AA

$△BDC\sim △ADB$

Hence $\frac{BD}{CD}=\frac{AD}{BD}$

$\frac{8}{CD}=\frac{4}{8}$

$CD=16$

Hence $CD=16cm$