How to prove the formula of altitude from this following triangle?

Question

Given: Right triangle $\triangle ABC$ with $A$ as right angle.

If $t_A$ is altitude that drawn from point $A$ to $\overline{BC}$, called $\overline{AD}$.

Prove that $t_A = \sqrt{2}\cdot\dfrac{bc}{b+c}$

Thanks

Answer 1

Taking @kmitov's suggestion that $\overline{AD}$ should be the bisector of $\angle A$, we have that $\overline{AD}$ is the diagonal of a square:

Thus,

$$\triangle BDF \sim \triangle DCE \; \implies \; \frac{s}{c-s} = \frac{b-s}{s} \; \implies \; \frac{b c}{b+c} = s = \frac{t}{\sqrt{2}} \; \implies \; t = \sqrt{2}\frac{bc}{b+c}$$

Answer 2

By the picture that you drew, we have $\frac{AD}{c}=\frac{b}{\sqrt{b^2+c^2}}$, which gives $AD=\frac{bc}{\sqrt{b^2+c^2}}$.

This is hardly ever equal to $\sqrt{2}\frac{bc}{b+c}$. For if we had equality, we would have $\frac{\sqrt{2}}{b+c}=\frac{1}{\sqrt{b^2+c^2}}$, or equivalently $2(b^2+c^2)=(b+c)^2$. This simplifies to $b^2-2bc+c^2=0$, that is, $b=c$.

So the proposed formula only gives the right answer if our triangle is right-angled and isosceles.