<s>[INST] How to prove the formula of altitude from this following triangle? 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
 [/INST] 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. 
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