Question

In given figure, if $\angle ACB=\angle CDA,AC=8cm$ and $AD=3cm$, then find BD.

Answer 1

Given, $AC=8cm,AD=3cm$ and $\angle ACB=\angle CDA$
From figure, $\angle CDA={90}^{\circ }$
$\therefore$ $\angle CDA={90}^{\circ }$


In right angled $\Delta ADC$, $A{C}^2=A{D}^2+C{D}^2$
$\Rightarrow$ $(8{)}^2=(3{)}^2+(CD{)}^2$
$\Rightarrow$ $64-9=C{D}^2$
$\Rightarrow$ $CD=\sqrt{55}cm$
In $\Delta CDB$ and $\Delta ADC$, $\angle BDC=\angle ADC$ $\left[each{90}^{\circ }\right]$
$\angle DBC=\angle DCA$ [each equal to ${90}^{\circ }-\angle A$]
$\therefore$ $\Delta CDB\sim \Delta ADC$
Then, $\frac{CD}{BD}=\frac{AD}{CD}$
$\Rightarrow$ $C{D}^2=AD\times BD$
$\therefore$ $BD=\frac{C{D}^2}{AD}=\frac{(\sqrt{55}{)}^2}{3}=\frac{55}{3}cm$