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In the adjoining figure, $\Delta CAB$ is a right triangle, right angled at $A$ and $AD \perp BC$. Prove that $\Delta ADB \sim \Delta CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of $AD$.
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$\Delta ABC \sim \Delta DAC$ (1 mark). Similarly, $\Delta ABC \sim \Delta DBA$ ($\frac{1}{2}$ mark). From equations ① and ②, $\Delta DAC \sim \Delta DBA$ or $\Delta ADB \sim \Delta CDA$ (1 mark). $\frac{AD}{CD} = \frac{BD}{AD}$ ($\frac{1}{2}$ mark). $AD^2 = BD \times CD = 8 \times 2$ ($\frac{1}{2} + 1$ marks). $\therefore AD = 4$ cm ($\frac{1}{2}$ mark).