Let's have a polygon. Let's draw a line at some angle $\theta_1$ to some fixed direction. By moving the line parallel to itself, we can make the whole polygon to lie either on one side of the line or the other. Thus, since the function “area on one side minus area on the another” is continuous, it should go through zero. Thus, for every $\theta_1$ there is a line $AD = L(\theta_1)$ that splits the polygon in halves.
Using the same mean value theorem we can show, that for every $\theta_1$, there is $\theta_2$ such that lines $L_1=AD=L(\theta_1)$ and $L_2=BE=L(\theta_2)$ split the area in a proportion $k=[APB]/[BPD]=1/2$. Indeed, if $\theta_2=\theta_1$, then $L_1=L_2$ and $k=0$. However, if $\theta_2=\theta_1+\pi$ (we rotated the line 2 by $\pi$ and $L_1=L_2$ again), then $k=\infty$. Given $k(\theta_2)$ is continuous, there should be some $\theta_2$, so $k=1/2$
By the same argument, for every $\theta_1$ there is always a $\theta_3$ and line $L_3=CF=L(\theta_3)$, so $k=[AQC]/[CQD]=2$.
Now consider point $R$, the intersection of $L_2$ and $L_3$. As we traverse from $\theta_1\to\theta_1+\pi$, lines $L2\leftrightarrow L3$ and $P\leftrightarrow Q$, but $R\to R$. However, now $R$ lies on the other side of line $L_1$. That means that during its journey, it crossed the line $L_1$. At this moment all three lines pass through one point.
By the way, we didn't use neither property of convexity, nor that it is a polygon. What mattered is that the area of the intersection of this shape with half-plane is continuous in respect to movements of the half-plane.