|AB|=|BC|=|CA|=2
P is on BC
let
|BP|/|PC|=k
AP = (AB+k(AC))/(1+k)
AP*(AB+AC)
=[1/(1+k)](AB+k(AC)).(AB+AC)
=[1/(1+k)]( |AB|^2 + k|AC|^2 + (k+1)AB.AC)
=[1/(1+k)]( 4 + 4k + 2(k+1))
=[1/(1+k)](6k+6)
=6