Question

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

Answer 1

In $\Delta BAC$,

$B{C}^2=A{C}^2+A{B}^2$

$B{C}^2=b^2+c^2$

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

Now, in $\Delta ABD$ and $\Delta CBA$,

$\angle B=\angle B$ (common)

$\angle ADB=\angle BAC$ (each ${90}^{\circ }$)

So, by AA similarity,

$\Delta ABD\sim \Delta CBA$

Since the corresponding parts of similar triangles are similar, then,

$\frac{AB}{CB}=\frac{AD}{AC}$

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

$AD=\frac{bc}{\sqrt{b^2+c^2}}$