Question on tangent lines and the center of an ellipse 
Let $\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$ be an ellipse. Let $A = (x_0,y_0)$ be a point outside of an ellipse. Draw two lines that touch an ellipse and denote two tangent points by $D_1 = (x_1,y_1),D_2 = (x_2,y_2)$. My question is that if I draw a line passing through $A$ and the middle point of $D_1$ and $D_2$ i.e., $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ then it passes the origin.


Here's the image I found on google. I think it's correct but I don't know how to prove this. I already know the equation of a line passing $D_1$ and $D_2$ is $\frac{xx_0}{a^2}+\frac{yy_0}{b^2} =1$. So what I need to show is
$$\frac{y_0}{x_0}\left(\frac{x_1+x_2}{2}\right) = \frac{y_1+y_2}{2}$$
But I don't know how to get further. How can I do this? Is there any simple geometric proof of this?

The equation of the line $D_1 D_2$ which is chord of contact of the ellipse is
$$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=1~~~~~~~(1)$$
$D_1D_2$ is also a chord whose mid point is $(h,k)$, where $h=(x_1+x_2)/2, k=(y_1+y_2)/2$, we have
$$\frac{xh}{a^2}+\frac{yk}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}=p~~~~(2)$$
Compare (1) and (2) to get $x_0=ph$ and $y_0=kp$ so we get
$$\frac{x_1+x_2}{x_0}=\frac{y_1+y_2}{y_0}$$
