Ta có:
\(M\in AB\), \(N\in AC\) \(\Rightarrow MN\subset (ABC)\)
Trong tam giác \(ABC\) ta có:
\(\dfrac{AM}{AB}=\dfrac{AN}{AC}\) \(\Rightarrow MN\parallel BC\)
Ta có \(D ∈ (DBC) ∩ (DMN)\) và
\(\left\{ \begin{array}{l}BC \subset (DBC)\\MN \subset (DMN)\\BC\parallel MN\end{array} \right.\)
\(\Rightarrow (DBC)\cap (DMN)=Dx,\)
\(Dx\parallel BC\parallel MN\).
