Solution:-

From the figure, consider ∆ABC,

So, ∠A = 90^o

And AD ⊥ BC

∠BAC = 90^o

Then, ∠BAD + ∠DAC = 90^o … [equation (i)]

Now, consider ∆ADC

∠ADC = 90^o

So, ∠DCA + ∠DAC = 90^o … [equation (ii)]

From equation (i) and equation (ii)

We have,

∠BAD + ∠DAC = ∠DCA + ∠DAC

∠BAD = ∠DCA … [equation (iii)]

So, from ∆BDA and ∆ADC

∠BDA = ∠ADC … [both the angles are equal to 90^o]

∠BAD = ∠DCA … [from equation (iii)]

Therefore, ∆BDA ~ ∆ADC

BD/AD = AD/DC = AB/AC

Because, corresponding sides of similar triangles are proportional

BD/AD = AD/DC

By cross multiplication we get,

AD² = BD × CD

AD² = 2 × 8 = 16

AD = √16

AD = 4