Trong \(\left( BCD \right)\) gọi \(E=MN\cap CD\).

Trong \(\left( ACD \right)\) gọi \(Q=AD\cap PE\).

Khi đó thiết diện của hình chóp khi cắt bởi mặt phẳng \(\left( MNP \right)\) là tứ giác MNQP.

Áp dụng định lí Menelaus trong tam giác BCD ta có:

\(\frac{MB}{MC}.\frac{EC}{ED}.\frac{ND}{NB}=1\Rightarrow \frac{1}{2}.\frac{EC}{ED}.\frac{1}{2}=1\Leftrightarrow \frac{EC}{ED}=4\).

Áp dụng định lí Menelaus trong tam giác ACD ta có:

\(\frac{PA}{PC}.\frac{EC}{ED}.\frac{QD}{QA}=1\Rightarrow 1.4.\frac{QD}{QA}=1\Rightarrow \frac{QD}{QA}=\frac{1}{4}\)

Ta có: \({{V}_{ABMNQ}}={{V}_{ABMN}}+{{V}_{AMNP}}+{{V}_{ANPQ}}\)

+) \(\frac{{{S}_{BMN}}}{{{S}_{BCD}}}=\frac{BM}{BC}.\frac{BN}{BD}=\frac{1}{3}.\frac{2}{3}=\frac{2}{9}\Rightarrow \frac{{{V}_{ABMN}}}{{{V}_{ABCD}}}=\frac{2}{9}\)

+) \(\frac{{{V}_{AMNP}}}{{{V}_{AMNC}}}=\frac{AP}{AC}=\frac{1}{2}\Rightarrow {{V}_{AMNP}}=\frac{1}{2}{{V}_{AMNC}}\)

\(\frac{{{S}_{NMC}}}{{{S}_{DBC}}}=\frac{d\left( N;BC \right).MC}{d\left( D;BC \right).BC}=\frac{NB}{DB}.\frac{MC}{BC}=\frac{2}{3}.\frac{2}{3}=\frac{4}{9}\)

\(\Rightarrow \frac{{{V}_{AMNC}}}{{{V}_{ABCD}}}=\frac{4}{9}\Rightarrow {{V}_{AMNP}}=\frac{2}{9}{{V}_{ABCD}}\)

+) \(\frac{{{V}_{APQN}}}{{{V}_{ACDN}}}=\frac{AP}{AC}.\frac{AQ}{AD}=\frac{1}{2}.\frac{4}{5}=\frac{2}{5}\Rightarrow {{V}_{APQN}}=\frac{2}{5}{{V}_{ACDN}}\)

\(\frac{{{S}_{CND}}}{{{S}_{CBD}}}=\frac{DN}{DB}=\frac{1}{3}\Rightarrow \frac{{{V}_{ACDN}}}{{{V}_{ABCD}}}=\frac{1}{3}\Rightarrow {{V}_{APQN}}=\frac{2}{15}{{V}_{ABCD}}\)

\(\Rightarrow {{V}_{ABMNQ}}={{V}_{ABMN}}+{{V}_{AMNP}}+{{V}_{ANPQ}}=\frac{2}{9}{{V}_{ABCD}}+\frac{2}{9}{{V}_{ABCD}}+\frac{2}{15}{{V}_{ABCD}}=\frac{26}{45}{{V}_{ABCD}}\).

Gọi \({{V}_{1}}={{V}_{ABMNQ}},{{V}_{2}}\) là thể tích phần còn lại \(\Rightarrow \frac{{{V}_{1}}}{{{V}_{2}}}=\frac{26}{19}\).