Question
$\triangle\text{ABD}$ is a right triangle right-angled at $A$ and $\text{AC}\perp\text{BD}.$ Show that
$AD^2= BD × CD$
$AD^2= BD × CD$
✓
Answer
In $\triangle\text{DCA}\ \&\ \triangle\text{DAB}$
$\angle\text{DCA}=\angle\text{DAB} ($Both are equal to $90^\circ )$
$\angle\text{CDA}=\angle\text{ADB} ($Common angle$)$
$\angle\text{DAC}=\angle\text{DBA} ($Remaining angle$)$
$\triangle\text{DCA}\sim\triangle\text{DAB}(AAA$ property$)$
Therefore $\frac{\text{DC}}{\text{DA}}=\frac{\text{DA}}{\text{DB}}$
$\Rightarrow\text{AD}^2=\text{BD}\times\text{CD}$
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