[THOUGHT] To solve the problem, we start by assigning coordinates to the vertices of the tetrahedron $ABCD$ based on the given conditions. We place $C$ at the origin $(0,0,0)$, $D$ at $(4,0,0)$, and $H$ (the foot of the perpendicular from $A$ to $CD$) at $(2,0,0)$. Given $AH = 4$, $A$ is at $(2, 4, 0)$. For $B$, we use the conditions $BC = BD = 2\sqrt{5}$ and find $B$ at $(2, 2, 2\sqrt{3})$.

The centroid $G$ of triangle $ABH$ is calculated as the average of the coordinates of $A$, $B$, and $H$, resulting in $G\left(2, 2, \frac{2\sqrt{3}}{3}\right)$.

The sphere $S$ centered at $G$ with radius $\frac{2\sqrt{3}}{3}$ is tangent to the plane $ACD$ (the $xy$-plane). The locus $T$ of points $P$ on $S$ such that $\angle APG = \frac{\pi}{2}$ forms a circle. The radius of this circle is found to be 1 by solving the intersection of the sphere with the plane defined by the condition $(P - A) \cdot (P - G) = 0$.

To find the area of the orthogonal projection of $T$ onto the plane $ABC$, we calculate the angle between the planes of $T$ and $ABC$. The normal vectors of these planes are $(0, 2, -\frac{2\sqrt{3}}{3})$ and $(2\sqrt{3}, -\sqrt{3}, -1)$, respectively. The cosine of the angle between these normals is $\frac{1}{4}$, leading to a scaling factor of $\frac{1}{4}$ for the area.

Thus, the area of the projection is $\pi \times 1^2 \times \frac{1}{4} = \frac{\pi}{4}$, giving $k = 4$.

[/THOUGHT] [PYTHON]

Verification of the key steps

Coordinates of points

A = (2, 4, 0) B = (2, 2, 2 * (3**0.5)) C = (0, 0, 0) D = (4, 0, 0) H = (2, 0, 0)

Centroid G of triangle ABH

G = ( (A[0] + B[0] + H[0]) / 3, (A[1] + B[1] + H[1]) / 3, (A[2] + B[2] + H[2]) / 3 )

Radius of sphere S

radius_S = abs(G[2]) # Distance from G to plane ACD (z=0)

Normal vectors

n1 = (A[0] - G[0], A[1] - G[1], A[2] - G[2]) # Normal of plane T n2 = (2 * (30.5), - (30.5), -1) # Normal of plane ABC

Dot product of normals

dot_product = n1[0] * n2[0] + n1[1] * n2[1] + n1[2] * n2[2]

Magnitudes of normals

magnitude_n1 = (n1[0]**2 + n1[1]**2 + n1[2]**2)**0.5 magnitude_n2 = (n2[0]**2 + n2[1]**2 + n2[2]**2)**0.5

Cosine of the angle between the planes

cos_theta = abs(dot_product) / (magnitude_n1 * magnitude_n2)

Area of projection

area_projection = 3.141592653589793 * (1**2) * cos_theta

Since the problem states the area is π/k, we solve for k

k = 3.141592653589793 / area_projection

print(k) [/PYTHON]

[THOUGHT] The Python output confirms that $k = 4.0$. Thus, the area of the orthogonal projection of $T$ onto the plane $ABC$ is $\frac{\pi}{4}$, and the value of $k$ is: \boxed{4}