Angle bisector in a right angled triangle In a right angled triangle, the legs adjacent to the right angle are equal to $a$ and $b$. Prove that the length of the bisector (of the right angle) is equal to $$\frac{a\cdot b\cdot \sqrt{2}}{a+b}.$$
While approaching this question, I was very puzzled as to how I would end up with this expression. 
Additionally, I couldn't figure out where the $\sqrt{2}$ would come from, other than the sine or cosine of $45$ degrees (from the bisector). 

An elemenraty solution: In the following figure $|BC|=a,|CA|=b,|AB|=c$ and $[CD]$ angle bisector. Let's draw square $CEDF$ and $|CE|=x$. So, $|BE|=a-x$ and  $|CD|=x\sqrt2$. Now $\triangle ABC \sim \triangle DBE$ and $$\dfrac{b}{a}=\dfrac{x}{a-x} $$
Therefore $x=\dfrac{ab}{a+b}$ and $|CD|=\dfrac{ab\sqrt2}{a+b}.$

