104
D is a point on the side BC of $\triangle ABC$ such that $\angle CAB = \angle CDA$. Show that $CA^2 = CB \times CD$.
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$\triangle ADC \sim \triangle BAC$ (1 Mark)
$\frac{DC}{AC} = \frac{AC}{BC}$ (1/2 Mark)
$\Rightarrow AC^2 = DC \times BC$ or $CA^2 = CB \times CD$ (1/2 Mark)
$\frac{DC}{AC} = \frac{AC}{BC}$ (1/2 Mark)
$\Rightarrow AC^2 = DC \times BC$ or $CA^2 = CB \times CD$ (1/2 Mark)