Question
$\triangle\text{ABD}$ is a right triangle right-angled at A and $\text{AC}\perp\text{BD}.$ Show that
$AD^2 = BD \times CD$
$AD^2 = BD \times CD$
✓
Answer
In $\triangle\text{DCA}\ \&\ \triangle\text{DAB}$
$\angle\text{DCA}=\angle\text{DAB}$ (Both are equal to 90°)
$\angle\text{CDA}=\angle\text{ADB}$ (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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