Prove that the line $CQ$ passes through a fixed a point Given $A(3,0)$ and $B(6,0)$ are $2$ fixed points and $P(x,y)$ is a variable point. $AP$ and $BP$ meet the y axis at $C$ and $D$ respectively. The line $OP$, $O$ being the origin intersects the line $AD$ at $Q$. Prove that the line $CQ$ passes through a fixed point.
The problem looks quite nasty. Family of lines isn't helping much. Though, drawing the figure, it looks like it passes through the mid point of $AO$. Some hints. Thanks.
 A: Let's denote the coordinates of the point $P\;$ by $\;(x_P,y_P)$, suppose $x_P\neq 0$, then we have the following equations
\begin{align*}
\text{Line }AP:\quad y&=\frac{y_P}{x_P-3}(x-3)\\
\text{Line }BP:\quad y&=\frac{y_P}{x_P-6}(x-6)\\
\end{align*}
(It is supposed that all denominators are $\neq 0$) Then point $C$ and $D$ have coordinates $\left(0,\frac{3y_P}{3-x_P}\right)$ and $\left(0,\frac{6y_P}{6-x_P}\right)$ respectively. Also we have the following equations for the lines
$$\text{Line }OP:\quad y=\frac{y_P}{x_P}x$$
$$\text{Line }AD:\quad y=\frac{\frac{6y_P}{6-x_P}}{-3}(x-3)$$
Let's denote by $(x_Q,y_Q)$ the coordinates of the point $Q$, we have
\begin{align*}
\frac{y_P}{x_P}x_Q=y_Q&=\frac{2y_P}{x_P-6}(x_Q-3)\\
 (x_P-6)x_Q&=2x_P(x_Q-3)\\
x_Q&=\frac{6x_P}{x_P+6}
\end{align*}
Then, an equation for the line $CQ$ is 
$$y=\frac{\frac{6y_p}{x_P+6}-\frac{3y_P}{3-x_P}}{\frac{6x_P}{x_P+6}}x+\frac{3y_P}{3-x_P}\quad\text{or}\quad y=\frac{3y_P}{2(x_P-3)}x+\frac{3y_P}{3-x_P}$$
It is clear that $CQ$ passes through $(2,0)$.
A: For, ease of understanding let the coordinates of point $P$ be $(h, k) $ then the equation of line $AP$ passing through $A(3, 0)$ $$y-k=\frac{k-0}{h-3}(x-h)$$
setting, $x=0$ in above equation,we get $$y=\frac{-3k}{h-3}$$ Hence, $$C\equiv\left(0, \frac{-3k}{h-3}\right)$$ 
Similarly, the equation of line $BP$ passing through $A(6, 0)$ $$y-k=\frac{k-0}{h-6}(x-h)$$
setting, $x=0$ in above equation,we get $$y=\frac{-6k}{h-6}$$ Hence, $$D\equiv\left(0, \frac{-6k}{h-6}\right)$$ 
Now, equation of line OP: $$y-0=\frac{k-0}{h-0}(x-0)$$
$$y=\frac{k}{h}x$$  Equation of line AD: $$y-0=\frac{\frac{-6k}{h-6}-0}{0-3}(x-3)$$ $$y=\frac{2k}{h-6}(x-3)$$
Now, solving the equations of the lines OP & AD, we get the coordinates of the point $Q$ as follows  $$Q\equiv \left(\frac{6h}{h+6}, \frac{6k}{h+6}\right)$$
Now, the equation of the line CQ is given as $$y-\left(-\frac{3k}{h-3}\right)=\frac{\frac{6k}{h+6}-\left(-\frac{3k}{h-3}\right)}{\frac{6h}{h+6}-0}(x-0)$$  $$y=\frac{3kx}{2(h-3)}-\frac{3k}{h-3}$$
$$y=\frac{3k}{2(h-3)}(x-2)$$ It is obvious that the line: CQ passes through a fixed point $\color{blue}{(2, 0)}$
