tangent on parabola bisects angle I'd like to show that the tangent on a parabola bisects the angle $\alpha$ of the triangle $QPF$:
Now I have the following:


*

*PQ || FT   and   PF || QT (do I need to show this?

*Now I wanted to show that $|PQ|=|PF|$ and $|FT|=|PQ|$. With this I have $QTFP$ as a rhombus!


Would this be a good way to show the "tangent bisect"-property as written above or is there a better/easier way?
How should I proceed?
 A: You probably should be specific about what your elements are, i.e. what you assume as their definition and what you deduce as additional properties. The way I read your figure, $P$ is an arbitrary point on the parabola, $Q$ is its orthogonal projection to the directrix, $F$ is the focus of the parabola and $T$ is the point where the tangent intersects the axis of symmetry. Right?
If this assumption is correct, then $PQ\parallel FT$ is a direct consequence of the definition. $PF\parallel QT$, on the other hand, is far from clear and would require a proof.
$\lvert PQ\rvert=\lvert PF\rvert$ is a direct consequence of the definition of a parabola via focus and directrix: the parabola is the set of all points for which the distance from the focus equals the (perpendicular) distance from the directrix. If you have a different definition of what a parabola is, then you may want to establish this property as a theorem first.
On the whole, it's a bit tricky now to bring the tangent into this. So perhaps it's better to turn things around: define $T$ as the point which completes $FPQ$ to a rhombus, and then show that the resulting line $PT$ is in fact the tangent. To that effect, show that $P$ is the only point on that line which also lies on the conic. I don't have a good argument for this just now, either, but perhaps someone else will edit that in or write a complimentary answer. Your comment suggests you might have solved this part of the problem yourself already.
Once you have established that there is no other point of intersection along the line, you can show that it is a tangent by an algebraic argument. From the family of lines parallel to this, there are some which intersect the conic twice, and others which don't intersect it at all (or only at complex points). In between these two there is the limit case where the two intersections coincide. And via a limit process, that corresponds to the situation where a secant (defined by two distinct points) turns into a tangent (defined by a single point of multiplicity two). Note that this argument only holds for points which actually have multiplicity two: it will fail e.g. for vertical lines and a hyperbola which has a vertical asymptote. There you don't get points of intersection of multiplicity two, but instead one of the points of intersection would be at infinity, which can be captured nicely using projective geometry. So perhaps the argument should include the observation that only lines parallel to the axis of symmetry do intersect the parabola at infinity. How much of this needs to be shown depends on how rigorous you aim to be and what background you have.
