In the adjoining figure, find the length of AD in terms of b and c.

In right angle triangle ABC,

By pythagoras theorem,

BC² = AB² + AC²

BC² = c² + b²

BC = $\sqrt{b^2 + c^2}$.

From figure,

Area of △ABC = Area of △ABD + Area of △ADC

⇒ $\dfrac{1}{2}$ x AB x AC = $\dfrac{1}{2}$ x AD x BD + $\dfrac{1}{2}$ x AD x CD

⇒ $\dfrac{1}{2}$(AB.AC) = $\dfrac{1}{2}$(AD.BD + AD.CD)

⇒ AB.AC = AD.BD + AD.CD

⇒ AB.AC = AD(BD + CD)

⇒ AB.AC = AD.BC [∵ BD + CD = BC]

⇒ AD = $\dfrac{\text{AB.AC}}{\text{BC}}$

Putting values of AB, AC and BC in above equation we get,

AD = $\dfrac{c.b}{\sqrt{b^2 + c^2}}$

Hence, AD = $\dfrac{bc}{\sqrt{b^2 + c^2}}$.