Question
If $\angle\text{ACB}=\angle\text{CDA},$ $AC = 8cm$ and $AD = 3cm$, find $BD.$
✓
Answer
Given, $AC = 8cm, AD = 3cm$ and $\angle\text{ACB}=\angle\text{CDA}$
From figure, $\angle\text{CDA}=90^\circ$
In right angles $\triangle\text{ADC},$
$AC^2= AD^2 + CD^2$
$\Rightarrow (8)^2 = (3)^2 + (CD)^2$
$\Rightarrow 64 - 9 = CD^2$
$\Rightarrow\text{CD}=\sqrt{55}\text{cm}$
In $\triangle\text{CDA}$ and $\triangle\text{ADC},$
$\angle\text{BDC}=\angle\text{ADC}$ [each 90°]
$\angle\text{DBC}=\angle\text{DCA}$ $[\text{each equal to }90^\circ-\angle\text{A}]$
$\therefore\triangle\text{CDB}\sim\triangle\text{ADC}$
Then $\frac{\text{CD}}{\text{BD}}=\frac{\text{AD}}{\text{CD}}$
$\Rightarrow\text{CD}^2=\text{AD}\times\text{BD}$
$\therefore\text{BD}=\frac{(\text{CD})^2}{\text{AD}}$
$\text{BD}=\frac{\big(\sqrt{55}\big)^2}{3}$
$\text{BD}=\frac{55}{3}\text{cm}$
From figure, $\angle\text{CDA}=90^\circ$
In right angles $\triangle\text{ADC},$
$AC^2= AD^2 + CD^2$
$\Rightarrow (8)^2 = (3)^2 + (CD)^2$
$\Rightarrow 64 - 9 = CD^2$
$\Rightarrow\text{CD}=\sqrt{55}\text{cm}$
In $\triangle\text{CDA}$ and $\triangle\text{ADC},$
$\angle\text{BDC}=\angle\text{ADC}$ [each 90°]
$\angle\text{DBC}=\angle\text{DCA}$ $[\text{each equal to }90^\circ-\angle\text{A}]$
$\therefore\triangle\text{CDB}\sim\triangle\text{ADC}$
Then $\frac{\text{CD}}{\text{BD}}=\frac{\text{AD}}{\text{CD}}$
$\Rightarrow\text{CD}^2=\text{AD}\times\text{BD}$
$\therefore\text{BD}=\frac{(\text{CD})^2}{\text{AD}}$
$\text{BD}=\frac{\big(\sqrt{55}\big)^2}{3}$
$\text{BD}=\frac{55}{3}\text{cm}$
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