Geometric proof of this property of the ellipse I came across the following property of the ellipse:

The distance from a focus of an ellipse to any point on the ellipse is equal to $a(1-e \cos\theta)$. 
  Where the $a$ is the length of semi-major axis and $\theta$ is the eccentric angle of the point.

I can prove this with coordinate geometry but I want a pure geometric proof of it. Please help.


Here $O$ is the centre, and $F$ is a focus.
It is known that $OF=ae$,$PN=b\sin\theta$ and $OQ=a$. Apply Pythagoras Theorem in the $\triangle PNF$:
$$PF^2=PN^2+NF^2=PN^2+(OF-ON)^2=PN^2+(OF-OQ\cos\theta)^2$$
$$=b^2\sin^2\theta+(ae-a\cos\theta)^2=a^2\sin^2\theta(1-e^2)+a^2e^2+a^2\cos^2\theta-2a^2e\cos\theta$$
$$=a^2(\sin^2\theta-e^2\sin^2\theta+e^2+\cos^2\theta-2e\cos\theta)=a^2(1+e^2\cos^2\theta-2e\cos\theta)$$Hence we get $PF=a(1-e\cos\theta)$
