151. Given a tetrahedron \(ABCD\). Another tetrahedron \(A_1B_1C_1D_1\) is positioned such that its vertices \(A_1, B_1, C_1\), and \(D_1\) lie respectively in the planes \(BCD\), \(CDA\), \(DAB\), and \(ABC\), and the planes of its faces \(A_1B_1C_1\), \(B_1C_1D_1\), \(C_1D_1A_1\), and \(D_1A_1B_1\) contain respectively the vertices \(D\), \(A\), \(B\), and \(C\) of the tetrahedron \(ABCD\). It is also known that \(A_1\) coincides with the centroid of the triangle \(BCD\), and the lines \(BD_1\), \(CB_1\), and \(DC_1\) bisect the segments \(AC\), \(AD\), and \(AB\) respectively. Find the volume of the common part of these tetrahedrons if the volume of the tetrahedron \(ABCD\) is \(V\).
151. Let $K, L$ and $M$ be the midpoints of $AB, AC$ and $AD$ (Fig. 30). From the problem statement, it follows that the tetrahedron $A_{1} B_{1} C_{1} D_{1}$ is bounded by the planes $D K A_{1}, B L A_{1}, C M A_{1}$, and the plane passing through $A$ parallel to $BCD$. In this case, the vertices $B_{1}, C_{1}$, and $D_{1}$ are positioned such that the points $M, K$, and $L$ are the midpoints of $CB_{1}, DC_{1}$, and $BD_{1}$ (the points $B_{1}, C_{1}, D_{1}$ are not shown on the figure).

Now let $Q$ be the midpoint of $BC$, and $P$ be the point of intersection of $BL$ and $KQ$. To find the volume of the common part of the two pyramids $ABCD$ and $A_{1} B_{1} C_{1} D_{1}$, we need to subtract from the volume of pyramid $ABCD - V -$ the volumes of three pyramids equal to $DKBQ$ (each of them has a volume of $\left.\frac{1}{4} V\right)$, and add the volumes of three pyramids equal to $A_{1} BQP$. The volume of the latter pyramid is $\frac{1}{24} V$. Thus, the volume of the common part is $\frac{3}{8} V$.