If a point lies outside a triangle, show that there is a line through that point that does not intersect the triangle I'm trying to prove that if a point is outside a triangle, then there is a line through that point that does not intersect the triangle.
This fact seems obvious, but I couldn't come up with a satisfying proof. I tried going by contradiction (that if any line intersects the triangle,then the point is surrounded by points contained on the triangle's sides, so the point is inside the triangle) but I do not find it satisfying, mainly because a line could intersect the triangle "in the backside"(this would only be a problem if the point were to lie on one of the triangle's sides though), but it still doesn't feel rigorous enough. Another idea would be to consider parallels to the sides, but I couldn't finish this proof well enough. 
Any very clear and rigorous methods to prove this? 
 A: If you extend the sides of the triangle to divide the space around the triangle into 6 regions, the point must lie in one of the regions or on one of the extended lines. You can number each region and from that, identify which line (point in the region adjacent to the side of a triangle including the boundary lines) or two lines (point in the region forming a vertical angle with an angle of the triangle excluding the boundary lines) will be parallel to the triangle side(s) and will not intersect the triangle.The proof is that these parallel lines will never enter the regions containing the other 2 sides of the triangle and the definition of parallel covers the 3rd side.
A: If you are to be rigorous, there are some theorems we may use from geometry without proof.
One of them is the hyperplane separation theorem, which actually states that given two convex compact subsets in Euclidean space there is a hyperplane that separates them.
So all you need to verify this theorem is that the given sets are compact and convex. A point, of course is compact and convex. A triangle is compact convex since it's bounded, and it is the intersection of the (closed) sectors of the angles which form it, which are closed convex sets  if the angles are non-reflex(intersection of closed convex sets is closed convex).
Hence, the theorem applies and gives a hyperplane, a line for us, which actually runs between the point and triangle, so that the point lies on one side and the triangle on the other. You can figure out how to produce a line actually passing through the point from here. It turns out that the above result generalizes  to the Hahn Banach theorem in infinite dimensions.
The approach by Michael above is also great, and it uses the fact that there is a minimum distance between two disjoint compact sets. This generalizes to Hilbert spaces very well, it turns out, where you actually have a unique point minimizing the distance between a set and a point if the set is convex.
A: Suppose $T$ is a triangular region in the plane. Can you justify the following geometric facts?


*

*For any point $A$, there exists a point $B$ in $T$ such that the distance from $A$ to $B$ is a minimum among all points of $T$.

*The foot of the altitude of a right triangle $\triangle ABC$ to its hypotenuse $BC$ lies strictly between $B$ and $C$.

*$A$ is the point on a line $L$ that is closest to a point $B$ not on $L$ if $AB$ is perpendicular to $L$.

*$T$ is convex. That is, if $A$ and $B$ are points of $T$ then the entire line segment $AB$ lies in $T$.


If you can prove those lemmas, then we can proceed as follows:
By (1), there is a point $Q$ in $T$ that is closest to $P$, and we choose our line $L$ perpendicular to $PQ$. If $L$ contains a point $R$ of $T$ then $\triangle PQR$ is a right triangle, and by (2) the altitude from $P$ falls strictly between $Q$ and $R$, say at the point $S$. Since $PS$ is perpendicular to $QR$, $S$ is the closest point on $QR$ to $P$, by (3). Since $T$ is convex and contains $Q$ and $R$, $T$ must also contain $S$, by (4). But the $S$ is a point of $T$ that is closer to $P$ than $Q$, which contradicts our choice of $Q$ in the first place. Therefore $L$ does not intersect $T$.
