Properties of harmonic function

[Bridgeport]

Let$S \in \mathbb{R}$ be a square and u a continuous harmonic function on $\bar{S}$. Show that the average of u over the perimeter of S is equal to
the average of u over the union of two diagonals.

[Хорхе]

I'm too lazy to work out a good solution... We can assume that $S=[-1,1]^2$ and $u(x,y)=u(y,-x)$ (otherwise take the symmetrisation of it; the "mean-value" fact for the initial function will follow easily from the one for the symmetrised function). Denote $v$ the harmonic conjugate of $u$ (Of course, v. Integrate $f(x+iy)=u(x,y)+iv(x,y)$ along the contour consisting of diagonal $d$ going from $[-1,-1]$ to $[1,1]$, side $-a$ going from $[1,1]$ to $[1,-1]$ and side $-b$ going from $[1,-1]$ to $[-1,-1]$. (Let $mv_A(g)$ denote the mean value of a function $g$ on a line $A$.) We obtain $\frac 12 mv_d(u+v) = mv_a(u)+mv_b(v), \frac 12mv_d(u-v) = mv_a(-v)+mv_b(u), now add these and use that $v(x,y)=v(-y,x)$ (because $f(iz)$ is analytic). I think I wrote too much for this problem. I should stop here.

[Bridgeport]

Thank you very much for the reply. I almost understood what you wrote, except this piece:

We obtain $\frac 12 mv_d(u+v) = mv_a(u)+mv_b(v), \frac 12mv_d(u-v) = mv_a(-v)+mv_b(u),

How do we obtain the above? Comparing real and imaginary part in simple fasion does not help. ;-(