Show that 5 lines in a pentagon are concurrent Pentagon $ABCDE$ is inscribed in a circle. For any edge of $ABCDE$, we can draw the line perpendicular to that edge that contains the centroid of the remaining three vertices. Show that these 5 lines are concurrent.



I have drawn a basic diagram but nothing has come to my mind so far.  All help is appreciated.
Long hint: 

Let the coordinates of the vertices of given pentagon be $A_i=(x_i, y_i), 1\leq i\leq 5.$ Then, let us compute the equation of the line that is passing through the centroid of triangle formed by $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and is perpendicular to the side $A_4A_5$. Observe that the centroid of the said triangle has coordinates $(a, b) = (\dfrac{x_1+x_2+x_3}{3}, \dfrac{y_1+y_2+y_3}{3}).$

Now the equation of the line that contains the side $A_4A_5$ is $$Y = \dfrac{y_5-y_4}{x_5-x_4}X+\dfrac{y_4x_5-y_5x_4}{x_5-x_4}.$$ 
If we let the equation of the desired line be $Y=mX+n$, then it satisfies $b = am+n$ and $m\dfrac{y_5-y_4}{x_5-x_4}+1=0.$ This is a system of linear equation, so you can easily find $m,n$. 

Can you proceed from here? If not, let me know and I can finish the rest.
The pentagon is regular. Using analytical geometry, let's take the centre of the circle to be the origin ( this is also the centre of the pentagon). The angle subtended at the centre by each side of the pentagon is $72^\circ$ ( I leave it to the reader to figure this out). Without any loss of generalization, we can assume one vertex to lie on the y-axis. Hence, the vertices of the pentagon are :

$$
(r\cos\frac{\pi}{10}, r\sin\frac{\pi}{10})
\\
(r,0)
\\
(r\cos\frac{9\pi}{10}, r\sin\frac{9\pi}{10})
\\
(r\cos\frac{17\pi}{10}, r\sin\frac{17\pi}{10})
\\
(r\cos\frac{25\pi}{10}, r\sin\frac{25\pi}{10})
$$
Now... your proposition can be easily proved.