Three normals of a parabola with tangent angles $a$, $b$, $c$ meet at a point with angle $d$. Prove $d=a+b+c-\pi$ The polar equation of a parabola is given by
$$r=a \ {\csc^2 \dfrac {\theta} {2}}$$
If the normals at three points on the parabola
whose vectorial angle are $a,b,c,$ meet at a point whose vectorial angle is $d,$ prove that 
$$d=a+b+c-\pi$$
I tried doing this by polar equations and using the general equations of normals whose vectorial angles are given.
It is a challenge for me to solve it.
I would appreciate a eloborate proof.
 A: Rewrite the equation of the parabola $r=a \ {\csc^2 \dfrac {\theta} {2}}$ in the Cartesian coordinates
$$y^2= 4t(x+t)\tag 1$$
(Note the parameter $t$, instead of $a$, is used to avoid confusion with the angle $a$.) 
The tangent of the normal at any point $(x,y)$ on the parabola is  $\tan\theta=-\frac{y}{2t}$. Assume that the three normals meet at the point $(p,q)$. The following equation can be established by matching their tangents,
$$\frac{q-y}{p-x}=-\frac{y}{2t}\tag 2$$
Eliminate $x$ from (1) and (2) to get
$$y^3-4t(p-t)y-8t^2q=0\tag 3$$
Note that the cubic equation (3) does admit three normals, corresponding to the tangent angles $a$, $b$ and $c$. Then, according to the Vieta's formulas,
$$y_a \>y_b\>y_c = 8t^2q$$
$$y_a + y_b +y_c = 0$$
$$y_a \>y_b+y_b\>y_c + y_c\>y_a= -4t(p-t)q$$
and the tangents of the three normals are respectively,
$$\tan a  =-\frac{y_a}{2t},\>\>\>\>\>\tan b  =-\frac{y_b}{2t},\>\>\>\>\>\tan c  =-\frac{y_c}{2t}$$
Use above results to evaluate
$$\tan (a+b+c) 
= \frac{\tan a + \tan b +\tan c - \tan a \tan b\tan c}{1-(\tan a \tan b+\tan b\tan c + \tan c\tan a)}$$
$$= \frac{-\frac1{2t}(y_a + y_b +y_c) +\frac1{8t^3}y_a y_by_c }{1-\frac1{4t^2}(y_a y_b+y_by_c + y_cy_a)}
= \frac pq = \tan d$$
which leads to either $d=a + b+ c$, or $d+\pi=a + b+ c$. To determine the valid expression, check the special case of $(p,q)$ on the $x$-axis, where we know $a+b=\pi$ and $c=d = 0$. Thus, 
$$d = a + b+ c-\pi$$
