Question
$\triangle\text{ABD}$ is a right triangle right-angled at A and $\text{AC}\perp\text{BD}.$ Show that
$\frac{\text{AB}^2}{\text{AC}^2}=\frac{\text{BD}}{\text{DC}}$
$\frac{\text{AB}^2}{\text{AC}^2}=\frac{\text{BD}}{\text{DC}}$
✓
Answer
From part (i) $AB^2 = CB \times BD$
From part (ii) $AC^2 = DC \times BC$
Hence $\frac{\text{AB}^2}{\text{AC}^2}=\frac{\text{CB}\times\text{BD}}{\text{DC}\times\text{BC}}$
$\frac{\text{AB}^2}{\text{AC}^2}=\frac{\text{BD}}{\text{DC}}$
Hence proved.
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