We apply the condition for static equilibrium ($\sum \vec{F} = 0$) to the connection points (knots) C and A. By symmetry, the tensions in the two AC ropes are equal, as are the tensions in the two AB ropes.

1. Analysis of Knot C: The tension in the lowest rope directly supports the weight.

   $$T_{CG} = W$$

   The vertical components of the two tensions $T_{AC}$ must balance the weight $W$.

   $$\sum F_y = 2 T_{AC} \cos\beta - W = 0$$ $$T_{AC} = \frac{W}{2\cos\beta}$$

   The horizontal components of the $T_{AC}$ tensions cancel due to symmetry.

2. Analysis of Knot A: For knot A to be in equilibrium under the action of only two forces ($T_{AC}$ and $T_{AB}$), the net force must be zero. We analyze the force components. Vertical equilibrium:

   $$\sum F_y = T_{AB} \sin\alpha - T_{AC} \cos\beta = 0 \implies T_{AB} = T_{AC} \frac{\cos\beta}{\sin\alpha}$$

   Substituting the expression for $T_{AC}$:

   $$T_{AB} = \left(\frac{W}{2\cos\beta}\right) \frac{\cos\beta}{\sin\alpha} = \frac{W}{2\sin\alpha}$$

   Horizontal equilibrium requires $T_{AB}\cos\alpha - T_{AC}\sin\beta = 0$. This condition is only met if the angles are complementary ($\alpha + \beta = \pi/2$), which makes $\sin\beta = \cos\alpha$ and $\cos\beta = \sin\alpha$. This specific geometry ensures that knot A is in equilibrium without any additional forces.

   For the special case $\alpha + \beta = \pi/2$, we use $\sin\alpha = \cos\beta$. The vertical equilibrium equation at knot A simplifies:

   $$T_{AB} \sin\alpha - T_{AC} \sin\alpha = 0 \implies T_{AB} = T_{AC}$$

   Therefore, the tension $T_{AB}$ is the same as $T_{AC}$.