Solution
= Solution
Let $D,E,F$ be the midpoints of $AB,AC,BC$. Write
$$
a=\mathbb E_A T_{BC},\qquad
d=\mathbb E_D T_{BC}=\mathbb E_E T_{BC},\qquad
f=\mathbb E_F T_{BC}.
$$
<First-step analysis> for the <simple random walk> gives
$$
a=1+d,
$$
because $A$ has neighbours $D,E$;
$$
d=1+\frac{a+d+f}{4},
$$
because $D$ has neighbours $A,B,E,F$; and
$$
f=1+\frac d2,
$$
because $F$ has neighbours $B,C,D,E$. Solving,
$$
d=4,\qquad f=3,\qquad
\boxed{\mathbb E_A T_{BC}=a=5}.
$$