Half of differential tage voltage gain
Here think of the differential mode half-circuit as the small-signal equivalent of a common-emitter with no degeneration. The gain of such circuit is \$A_v= \frac{v_o}{v_i}=\$ \$-g_m.R_C\$. Here \$v_i=\frac{v_d}{2}\$ and \$g_m=\frac{\beta}{r_\pi}\$. Substituting \$v_i\$ in \$A_v\$ you get \$A_{dm_1}=\frac{2v_{o1}}{v_d}=\$ \$-g_m.R_c\$. Now dividing both sides by 2, you get \$A_{dm_1}=\frac{v_{o1}}{v_d}=\frac{-g_m.R_c}{2}\$. This is the differential gain of a single-ended output of a differential pair.