Magnetic Field of Triangle

What is the magnetic field at a distance $z$ from the center of an equilateral triangle of side $a$, that has a current $I$. Verify that it reduces to the field of a dipole with the appropiate dipole moment $z>>a$.

Triangles Current (Griffiths)

Figure is above
$H=\sqrt{(a^2/12+z^2)}$ and
$\sin⁡ α= (a/2)/\sqrt{(a^2/3+z^2)}=\sqrt{3/2} a/\sqrt{(a^2+3z^2)}$
B field at point P is (grey triangle)
$B(P)=μI/4πH*(\sin ⁡α+\sin⁡ α) =$
$=(μI*2√3)/(4π\sqrt{(a^2+12z^2 )}*2* (\sqrt{3}/2) (a/\sqrt{(a^2+3z^2 )})=$
$=3μI/(2π\sqrt{(a^2+12z^2 )(a^2+3z^2})$

In the pink triangle
$\sin⁡ β=(a*√3/6)/\sqrt{(z^2+a^2/12)}=(a√3)/\sqrt{(12z^2+a^2)}*(2√3)/6=a/\sqrt{(12z^2+a^2 )}$

$B_{tot}=3*B_z=3*B(P)*\sin⁡ β=$
$=3* 3μIa/(2π\sqrt{(a^2+12z^2 )(a^2+3z^2)})*a/\sqrt{(12z^2+a^2 )}=$
$=(9μIa^2)/2π(12z^2+a^2 ) *(1/\sqrt{(a^2+3z^2)})$
When z >>a
$B_{tot}=(9μIa^2)/2π*1/(12 √3*z^3 )=(√3 μIa^2)/(8πz^3)$

The field of a magnetic dipole at distance $r$ is
$B_{dipole}=μ_0/4π [(3r(m*r))/r^5 -m/r^3]=μ_0/4π*2m/z^3$
with $m=I*Area=I*(a*(a√3)/2)/2=I (a^2 √3)/4$
Therefore
$B_{dipole}=(μ_o Ia^2 √3)/(8πz^3)$