Question
$\triangle\text{ABD}$ is a right triangle right-angled at $A$ and $\text{AC}\perp\text{BD}.$ Show that
$AB^2= BC × BD$
$AB^2= BC × BD$
✓
Answer
In $\triangle\text{ADB}$ and $\triangle\text{CAB}$
$\angle\text{DAB}=\angle\text{ACB}=90^\circ$
$\angle\text{ABD}=\angle\text{CBA} ($Common angle$)$
$\angle\text{ADB}=\angle\text{CAB} ($remaining angle$)$
So, $\triangle\text{ADB}\sim\triangle\text{CAB} ($by $AAA$ similarity$)$
Therefore $\frac{\text{AB}}{\text{CB}}=\frac{\text{BD}}{\text{AB}}$
$\Rightarrow\text{AB}^2=\text{CB}\times\text{BD}$
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