Let $X = LK \cap BC$ and $Y = EF \cap BC$. Then, we know $(XD;BC)$ and $(YD;BC)$ are both harmonic bundles, so $X = Y$. This means that $KL, EF || BC$, which implies $BD = DC$ through Ceva.

Same as above. After finding out that $X=Y$, this is simply impossible, unless $X=Y=P_{\infty}$, hence by midpoints and parallel lines, it is clear that D is the midpoint of $BC$

Since $AD$, $BE$, and $CF$ are concurrent, then $(MD;BC)$ is harmonic. Similarly, we have $(ND;BC)$ harmonic. Then, since $-1=(MD;BC)=(ND;BC)$, then $M=N$. But this is only possible if $LK \parallel EF \parallel BC$, which means that $D$ must be the midpoint of $BC$. This means that the only possible answer is $\boxed{1}$.

The answer is 1 and 1 only. Observe that $\overline{EF}$ and $\overline{LK}$ concur at a point $X$ on $\overline{BC}$ such that $(BC; DX)=-1$. Then $(BC; DP_\infty)=-1$ in the parallel case, implying $D$ is the midpoint.

By Ceva-Menelaus clearly $(EF \cap BC, D; C, B) = (KL \cap BC, D; C, B) = -1$ which implies that $EF$, $KL$ and $BC$ concur at the point at infinity due to $KL \parallel EF$. However since $-1 = (B, C; D, \infty) \implies D$ is the midpoint of $BC$.

We claim the only possible ratio is $1$, let $\overline{LK} \cap \overline{AD} = G$, and $\overline{EF} \cap \overline{AD} = H$ note that by ceva menelaus $-1=(EF; H \infty)= (LK; G \infty)$ so $\overline{AD}$ bisects $\overline{LK}$, and $\overline{EF}$. Also note that $-1=(LK; G \infty) \stackrel{P}=(BC; D, \overline{\infty P} \cap \overline{BC})$, and $-1=(EF; H \infty) \stackrel{Q}= (BC; D, \overline{Q\infty} \cap \overline{BC})$. Since $\overline{Q \infty} \cap \overline{BC} = \overline{ P \infty} \cap \overline{BC}$, $\overline{BC} \parallel \overline{EF} \parallel \overline {KL}$. Thus we can conclude $D$ is the midpoint of $BC$.

Notice that \[(A,L;F,B) \overset{C}{=} (A,P;Q,D) \overset{B}{=} (A,K;E,C),\] so by Prism Lemma we have $KL \parallel EF \parallel BC$. Thuen Cevalaus gives \[-1 = (KJ \cap BC, D; B, C) = (\infty D; BC) \implies \boxed{\frac{BD}{DC} = 1}. \quad \blacksquare\]