βAB=1  ⟺  βnB′h=0\beta_{AB} = 1 \iff \pmb \beta_{\pmb nB}'\pmb h =0

1=βAB=Cov(rA,rB)Var(rB)=Cov(w′r,rB)Var(rB)=Cov(w1r1+⋯+wnrn,rB)Var(rB)=Cov(w1r1,rB)+⋯+Cov(wnrn,rB)Var(rB)=w1β1B+⋯+wnβnB=w′βnB=(h+b)′βnB=h′βnB+b′βnB=h′βnB+1  ⟹  h′βnB=0\begin{align} 1 &= \beta_{AB} \\[10pt] &= \frac{Cov(r_A, r_B)}{Var(r_B)} \\[10pt] &= \frac{Cov(\pmb w'\pmb r, r_B)}{Var(r_B)} \\[10pt] &= \frac{Cov(w_1 r_1 + \cdots + w_n r_n, r_B)}{Var(r_B)} \\[10pt] &= \frac{Cov(w_1 r_1, r_B) + \cdots + Cov(w_n r_n, r_B)}{Var(r_B)} \\[10pt] &= w_1\beta_{1B} + \cdots + w_n\beta_{nB} \\[5pt] &= \pmb w' \pmb \beta_{\pmb nB} \\[5pt] &= (\pmb h + \pmb b)' \pmb \beta_{\pmb nB} \\[5pt] &= \pmb h'\pmb \beta_{\pmb nB} + \pmb b'\pmb \beta_{\pmb nB} \\[5pt] &= \pmb h'\pmb \beta_{\pmb nB} + 1 \\ \implies \pmb h'\pmb \beta_{\pmb nB} &= 0 \end{align}