Kindly solve the following.

Solution :
The solution/s to your 13th question has been provided below:


$$\begin{gathered} \mathrm{In} ∆ \mathrm{ABC}, \mathrm{we} \mathrm{have}, \angle \mathrm{A} = 90°. \\ \mathrm{Also}, \mathrm{AD}\perp \mathrm{BC}. \\ \mathrm{Now}, \\ \mathrm{In} ∆\mathrm{ABC}, \mathrm{we} \mathrm{have}, \\ \angle \mathrm{BAC} = 90° \\ \Rightarrow \angle \mathrm{BAD} + \angle \mathrm{DAC} = 90° .....\left(1\right) \\ \mathrm{In} ∆\mathrm{ADC}, \mathrm{we} \mathrm{have}, \\ \angle \mathrm{ADC} = 90° \\ \mathrm{So}, \angle \mathrm{DCA} + \angle \mathrm{DAC} = 90° ......\left(2\right) \\ \mathrm{from} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{have} \\ \angle \mathrm{BAD} + \angle \mathrm{DAC} = \angle \mathrm{DCA} + \angle \mathrm{DAC} \\ \Rightarrow \angle \mathrm{BAD} = \angle \mathrm{DCA} ......\left(3\right) \\ \mathrm{In} ∆\mathrm{BDA} \mathrm{and} ∆\mathrm{ADC}, \\ \angle \mathrm{BDA} = \angle \mathrm{ADC} \left(90° \mathrm{each}\right) \\ \angle \mathrm{BAD} = \angle \mathrm{DCA} \left(\mathrm{proved} \mathrm{above}\right) \\ \mathrm{so}, ∆\mathrm{BDA} ~ ∆\mathrm{ADC} \left(\mathrm{AA} \mathrm{similarity}\right) \\ \Rightarrow \frac{\mathrm{BD}}{\mathrm{AD}} = \frac{\mathrm{AD}}{\mathrm{DC}} = \frac{\mathrm{AB}}{\mathrm{AC}} \left(\mathrm{Corresponding} \mathrm{sides} \mathrm{of} \mathrm{similar} ∆\text{'}\mathrm{s} \mathrm{are} \mathrm{proportional}\right) \\ \Rightarrow \frac{\mathrm{BD}}{\mathrm{AD}} = \frac{\mathrm{AD}}{\mathrm{DC}} \\ \Rightarrow {\mathrm{AD}}^2 = \mathrm{BD} \times \mathrm{CD} \\ \Rightarrow {\mathrm{AD}}^2 = 2 \times 8 = 16 \\ \Rightarrow \mathrm{AD} = 4 \mathrm{cm} \end{gathered}$$
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