Splitting the plane to fit convexes I'm trying to show the following :
Let $K,L$ two closed convexes of $\mathbb{R}^2,O=(0,0)$


*

*If $O\notin K$ then there exists a straight line $D$ going through $O$ such that $K$ is in one of the half of plane defined by $D$

*If $K$ is bounded and does not intersect with $L$ then there exists a straight line $D$ such that $K$,$L$ are in two distinct halves of plan defined by $D$

Those two properties are really easy to understand intuitively or with a graph, but I haven't been able to find a proper mathematical proof.
 A: Two hints for the price of one: For the first one, consider the point $P\in K$ closest to the origin, and take $D$ to be normal to the line from $O$ to $P$.
For the second one, apply the first one to the set $K-L=\{u-v\colon u\in K, v\in L\}$.
A: Here's a solution to the first problem. (For these sorts of problems I suggest drawing pictures - it's amazing how helpful blobs on a page can be.) Harald's gives you the second.
Because $K$ is closed, there's a point in $K$ of minimal distance to the origin; call it $v=(x,y)$. Then consider the normal line $D = \text{span}(-y,x)$. It suffices to show that $K \cap D = \varnothing$. Suppose not; say $(-ty,tx) \in K$. Because $v$ had minimal distance to the origin, $t \geq 1$. By convexity, every point of the form $s(-ty,tx) + (1-s)(x,y) \in K$ for $0 \leq s \leq 1$. Let's take the norm of this point. After some algebra, we obtain $$\sqrt{(1-s)^2+s^2t^2}\|v\|.$$
But for $s$ sufficiently small (pick $0 < s < \frac{2}{1+t}$), the first term is less than 1; this contradicts that $v$ was the point of $K$ of least norm. So $K \cap D = \varnothing$ as desired.
