parabola given a fixed point for the focus and a line (segment) for the directrix.

I started with the focus, and directrix and a random point, labeled A, not the segment endpoint on the directrix. I created a segment connecting the focus and point A. I found the midpoint of this segment and constructed the perpindicular to the segment through the midpoint to construct the perpindicular bisector, which we will label "perpbisect". This is a tangent line to the parabola.

Next to find the point that is on the parabola, we know that the point must be the same distance from the focus as from the point on the directrix. To find this point, we draw the perpindicular to the directrix through the random point on the directrix. The point of intersection of the perpindicular through the midpoint and the perpindicular to the directrix through our random point will be the point on the parabola.

Now I attempt to give a geometric account of why this is true. A parabola is made up of all points equdistant from the focus and the directrix. To show that the point of intersection is on the parabola, we need to show that the distance from the point of intersecection to the focus and to the directrix are equal. If we connect the point of intersection to the focus with a line segment, then we can define two triangles, one with vertices at the intersection point, the focus, and the midpoint, and the other with vertices at the intersection point, random point and midpoint.

These two triangles are congruent by SAS. The midpoint divides the first segment between the focus and random point into two equal segments. The perpindicular line throught the midpoint is common to both triangles, and then the angle in between the two segments are equal right angles. Therefore, the segment from the intersection point to the focus has equal measure to the segment from the intersection point to the random point and the intersection point is equdistant from the directrix and the focus point. This shows that the intersection point is on the parabola.