Question
$\triangle\text{ABD}$ is a right triangle right-angled at A and $\text{AC}\perp\text{BD}.$ Show that
$AC^2 = BC \times DC$
$AC^2 = BC \times DC$
✓
Answer
Let $\angle\text{CAB}=\text{x}$
In $\triangle\text{CBA}$
$\angle\text{CBA}=180^\circ-90^\circ-\text{x}$
$\angle\text{CBA}=90^\circ-\text{x}$
Similarly in $\triangle\text{CAD}$
$\angle\text{CAD}=90^\circ-\angle\text{CAD}=90^\circ-\text{x}$
$\angle\text{CDA}=90^\circ-\angle\text{CAB}$
$=90^\circ-\text{x}$
$\angle\text{CDA}=180^\circ-90^\circ-(90^\circ-\text{x})$
$\angle\text{CDA}=\text{x}$
Now in $\triangle\text{CBA}$ and $\triangle\text{CAD}$ we may observe that,
$\angle\text{CBA}=\angle\text{CAD}$
$\angle\text{CAB}=\angle\text{CDA}$
$\angle\text{ACB}=\angle\text{DCA}=90^\circ$
Therefore $\triangle\text{CBA}\sim\triangle\text{CAD}$ (by AAA rule)
Therefore $\frac{\text{AC}}{\text{DC}}=\frac{\text{BC}}{\text{AC}}$
$\Rightarrow\text{AC}^2=\text{DC}\times\text{BC}$
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