How to show this fraction is equal to 1/2?

$$ \frac{(2(\frac {a}{\sqrt{2}})+a)a} {2(1+\sqrt{2})a^2}= \frac{(\sqrt{2}a+a)a} {2(1+\sqrt{2})a^2}= \frac{(\sqrt{2}+1)a^2} {2(1+\sqrt{2})a^2}={1\over2}. $$

I think you may have missed that by definition,

$\sqrt{2}\sqrt{2}=2$

And thus,

$\frac{2}{\sqrt{2}}=\sqrt{2}$

This simplification issue is quite common. Of course using this and multiplying out/ factoring terms may get your desired result:

$$=\frac{(\sqrt{2}+1)a^2}{2(\sqrt{2}+1)a^2}$$

In which the $\frac{(\sqrt{2}+1)a^2}{(\sqrt{2}+1)a^2}$ reduces to one in the case $a \neq 0$ .

Assume $a\neq 0$, we have \begin{align*} \frac{\left(2\left(\frac {a}{\sqrt{2}}\right)+a\right)a} {2(1+\sqrt{2})a^2}&=\frac{\left(\frac{2a}{\sqrt 2}+a\right)}{2(1+\sqrt 2)a}\tag 1\\ &=\frac{\left(\frac{2}{\sqrt2}+1\right)a}{2(1+\sqrt 2)a}\tag2\\ &=\frac{\left(\frac{2}{\sqrt2}+1\right)}{(2+2\sqrt 2)}\\ &=\frac{\sqrt 2}{\sqrt 2}\cdot \frac{\left(\frac{2}{\sqrt2}+1\right)}{(2+2\sqrt 2)}\\ &=\frac{2+\sqrt 2}{2\sqrt 2+2\cdot 2}\\ &=\frac{2+\sqrt 2}{2\sqrt 2+4}\\ &=\frac{(2+\sqrt 2)}{2(2+\sqrt 2)}\\ &=\frac{1}{2} \end{align*} where in $(1)$ I cancelled the first $a$ and in $(2)$ I factored out the second $a$ to cancel it also.