Proving that a line which passes through a point inside an angle always intersects one of its sides How can I prove that a line which passes through a point inside an angle $\alpha (0^{°} < \alpha < 180^°$) always intersects one of its sides? (By the word "side" I mean that the sides of the angle are rays) 
If our line is parallel to one of its sides, then it should intersect the other side. I proved this by contradiction. Then I tried to solve the problem with the help of coordinates, where I have the sides of the angle $y=0$ and $y=kx$ and the line $y=mx+n$. In order to finish this proof, I need to consider too many cases (with $k > 0$ and $k < 0$, $m > 0$ and $m < 0$, $n > 0$, $n < 0$). Is there an easier way to prove this obvious fact?

 A: Trying to calculate everything gives a lot of different cases. But essentially every step is the same. I don't know if you are familiar with affine geometry, but that way you can show it in one step.
Since $0° < \alpha < 180°$, there exists an affine transformation that maps the halflines to the positive $x$- and $y$-axis. Affine transformations preserve incidences (i.e. intersections, relative positions, etc.) so the point $P$ lies in the first quadrant. It suffices to prove this single case.
Edit: the part below was given before it was mentioned that $AB$ and $AC$ are rays (halflines) and that the angle should be less than $180$°.
I think the statement isn't complemently true. 
If the angle between the two sides is $180$ or $360$ degrees, then the sides $AB$ and $AC$ form one line segment. If you take a point not one $AB$ of $AC$, and draw a parallel line through it, there will be no intersection.
Another problem is: the answer depends on the lengths of your sides and what you mean with "being inside the angle". For instance, take the point $(7,2)$ in your picture and take a line through it parallel with $AB$. Then it will not cut the side $AC$.
Problem: the side $AC$ is "too short" or the point $(7,2)$ doesn't "lie inside the angle ABC properly enough". (I know, it's a strange formulation, but I hope you know what I mean.)
One thing is sure: if you use full lines instead of sides, and if the angle is not a multiple of 180 degrees, there is an intersection, because a line cannot be parallel to two other lines that are not parallel.
A: Your line is $y = mx + n$.  And your point is ... lets call it $(a,b)$.  Consider the line from $(0,0)$ to $(a,b)$.  That line is $y = \frac ba x$.
If we compare the slopes... well actually the angles of the slopes... $\arctan \frac ab$ to $\arctan m$. (You don't have to do the math; we want to know if just want to know how your line compares to the line through the origin).
Compare $\arctan \frac ba$ to $\arctan m$.  There are three cases:
Case 1: $\arctan \frac ab < m$.  In which case the line through the origin is "steeper" (or shallower in the negative direction) then your line.  And therefore the line will intersect the $x$ axis at a positive $x$ point.  (i.e. the $x$-axis ray).
Case 2: $\arctan \frac ab = \arctan m$.  In which case you line passes through the origin and intersects both rays.
Case 3: $\arctan \frac ab > \arctan m$.  Either your line is parallel to the $x$ axis or intersects the $x$ axis at a negative $x$ value.  Either way it will intersect your $y = mk + n$ line.
=== older answer ====
Euclid's Prop 1. 16 https://mathcs.clarku.edu/~djoyce/java/elements/bookI/propI16.html 
Orient you plane so that the one of the rays is horizontal, your point is "above" that ray and the other ray is above your point.  Draw a line from your point to the vertex of the angle.  Compare the slope of this line to the slope of your original line.  
If the slope is equal than the lines are the same and the line intersects both rays at the vertex.
If the slope is greater/steeper then your line then intersects the horizontal line to the "right" of the vertex off the ray.
If the slope is less/shallower then your line intersects the horizontal line to the "left" of the vertex "off" the ray.  However as your point is to the right and above the horizon, the other ray is above the horizon and to the left of your point, and your line intersects at the horizon to the left of the vertex and to the left of the other ray, it must intersect the other ray.
==== oldest answer ====
Euclids Fifth postulate:
The line through a point is either parallel to both sides, parallel to one side but not the other, or parallel to neither side.
If it is parallel to both sides then the two sides are a parallel to each other.  But that's a contradiction because the two sides meet at a vertex.  
And the other three all state the line intersects one of its sides.
Oh, wait.. you mean why if it will intersect the line on the same side of the vertex and not the other.  Well, that is because you is "inside" the angle.  
It must pass through the other line to get point on the other side of the vertex.
