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 $(i) A B^2=C B \times B D$
From part $(ii) A C^2=D C \times B C$
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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