Question: Prove that if A is convex set and $\alpha, \beta ≥ 0$ then $(\alpha + \beta)A = \alpha A + \beta A$

What came first on my mind is that I have to show that $(\alpha + \beta)A\subset \alpha A + \beta A$ and $\alpha A + \beta A\subset (\alpha + \beta)A$. But I don't know how to use convexity of A to show this.

Answer: 
One direction doesn't require convexity. Let $(\alpha+\beta)a\in (\alpha+\beta)A$. Then $\alpha a\in \alpha A$ and $\beta a\in \beta A$, and $(\alpha+\beta)a=\alpha a+\beta a$, so $(\alpha+\beta)a\in \alpha A + \beta A$.

For the other direction, let $\alpha a\_1 + \beta a\_2\in \alpha A + \beta A$. We need to find some $a\_3\in A$ such that $\alpha a\_1+\beta a\_2=(\alpha+\beta)a\_3$. To find such an $a\_3$, divide both sides by $(\alpha+\beta)$. (footnote\*). We set $a\_3=\frac{\alpha}{\alpha+\beta}a\_1+\frac{\beta}{\alpha+\beta}a\_2$. Because $A$ is convex and these fractions are in $(0,1)$ and add up to 1, $a\_3\in A$ and we're done.

(footnote\*) This proof frays at the edges if $\alpha=\beta=0$, so this special case must be handle separately; it's easy though since $(0+0)A=0A+0A=\{0\}$ for any $A$.
For the open half-space (I suppose in an n-vector space over the real numbers), consider that up to an isomorphism they're $\{ (x\_1,...,x\_n): x\_n >0\} $ so that each convex combination remains into that set.