how does one divide stress in such a bridge? I want to build a spaghetti bridge like this:

I have two spaghetti types: T-type (designed to tolerate tension) and C-type (designed to tolerate compression).
A mass $m$ is hanging from the point shown in the picture.
How can I calculate the tension of the spaghetties to know whether the bridge will break or not?
Angles are 36° (180°/5).
I have written the forces for the joints. Assume that the joints and spaghetties are massless.
I haven't written any forces for the two side joints because they are support points and thus connected to fixed object so all the forces on them cancel.
The spaghetties which are being stretched are T-type and the other ones are C-type.
I know the tension of T spaghetties makes C-types' compression. But I have trouble calculating. Help me.
My own approach was:
At the hanging point we have 4 tensions and each are:
$$\vec{T_1}=T_1\cos36\hat i+T_1\sin36\hat j$$
$$\vec{T_2}=T_2\cos72\hat i+T_2\sin72\hat j$$
$$\vec{T_3}=-T_3\cos72\hat i+T_3\sin72\hat j$$
$$\vec{T_4}=-T_4\cos36\hat i+T_4\sin36\hat j$$
Also a gravity force:
$$\vec{F_g}=-mg\hat j$$
The sum of all of them is zero because the point doesn't move. so:
$$(T_1\cos36 +T_2\cos72 -T_3\cos72 -T_4\cos36)\hat i + (T_1\sin36 +T_2\sin72 +T_3\sin72 +T_4\sin36 -mg)\hat j=0$$
Thus:
$$T_1\cos36 +T_2\cos72 -T_3\cos72 -T_4\cos36 =0$$
$$T_1\sin36 +T_2\sin72 +T_3\sin72 +T_4\sin36 =mg$$
At this point I'm stuck.
Can anyone please provide me some hint so that I can proceed? I'm not wanting for any solution or do-this-for-me request; just a bit hint.
 A: a) The easy way is to use a computer simulation program such as Force Effect (free)


b) The more appropriate way is to draw free body diagrams and calculate the tensions from the equilibrium equations

Equilibrium on the support B
$$ \left. \begin{aligned} 2 A_y & = m g \\ A_x & = C \cos 72° \\ A_y & = C \sin 72° \end{aligned} 
\right\} \begin{aligned} A_y &= 0.5 m g \\ A_x &= 0.162 m g \\ C & = 0.5257 m g \end{aligned}$$
Equilibrium on the node D
$$ \left. \begin{aligned} C \cos 72° + T \cos 36° - C \cos 36° & =0
\\ C \sin 72° - T \sin 36° - C \sin 36° & = 0 \end{aligned} \right\} \begin{aligned}
T & = 0.618 C = 0.325 m g \end{aligned} $$
A: I assume that the right and left corner are used to support the bridge. By assuming that the bridge is in static equilibrium you can find the external forces at the support points. Namely the sum of all external forces should add up to zero. Similar for the torque around each point. This should be enough to find the external forces at the supports, which then can used to find the tension/compression in each spaghetti stick, by using that the sum of all forces in each node should add up to zero.
