Prove that in a parabola the tangent at one end of a focal chord is parallel to the normal at the other end.
Now, I know prove this algebraically, and that's very easy, but I am not getting any visual picture of the above situation. It'd be great if someone could give a proof without (or with minimal) words for this one - these proofs are exciting!
EDIT: A quick-n-dirty working of what I call an algebraic proof:
WLOG, let the equation of the parabola be $y^2 = 4ax$. The coordinates of points on the focal chord and also on the parabola are: $ P(at^2,2at) $ and $Q(al^2, 2am)$. For these points to lie on a focal chord, $ tm = -1 $. The tangent at $P$ is given as $ y(2at) = 2a(x+at^2) $ so the slope is $1/t$. Similarly the slope of tangent at $Q$ is $1/m$. So the slope of the normal at $Q$ is $-m = 1/t = $ slope of tangent at $P$.
Now, I know I am using some 'shortcuts' here, but then this was just a fast-paced look at what I did. The point is, I know how to prove the statement using the regular 'equations approach'. I want to know if there are any visual proofs.