Connected diode transistor collector current
\$I_c\$ is always \$\beta \cdot I_b\$ in this simple model. Your idea of a "short circuit" between collector and base never exists for a BJT, because its function doesn't rely on voltages but currents.







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So, it looks like a "short circuit", but it isn't because the current distribution between the \$I_b\$ and \$I_c\$ paths is controlled by the current sources.

$$I_c = \beta \cdot I_b$$
$$\beta = \mathrm{const}$$
$$I_1 = I_c + I_b$$

$$\Rightarrow I_1 = \beta \cdot I_b + I_b = (1 + \beta) \cdot I_b$$
$$\Rightarrow I_b = \frac{1}{1+\beta} \cdot I_1$$
$$\Rightarrow I_c = \frac{\beta}{1+\beta} \cdot I_1$$

Set \$\beta = 1\$ in your thoughts. Base and collector path have identical properties now. You see how the fixed current \$I_1\$ splits into two equal parts.

To find out \$U_{be}\$, one has to look up this reduced Ebers-Moll model:

$$I_c = I_S \cdot e^\frac{U_{be}}{U_T}$$
$$I_S = \mathrm{const} ; U_T = \mathrm{const} $$

So

$$\frac{\beta}{1+\beta} \cdot I_1 = I_S \cdot e^\frac{U_{be}}{U_T}$$
$$\Rightarrow U_{be} = \ln(\frac{\beta}{1+\beta} \cdot \frac{I_1}{I_S}) \cdot U_T $$