In tetrahedron $ ABCD $, vertex $ D $ is connected with $ D_{0} $, the centrod in $ \triangle ABC $. Line parallel to $ DD_{0} $ are drawn through $ A,B $ and $C$. These lines intersect the planes $ BCD, CAD $ and $ ABD $ in points $ A_{1}, B_{1} $ and $ C_{1} $, respectively. Prove that the volume of $ ABCD $ is one third the volume of $ A_{1}B_{1}C_{1}D_{0} $. Is the result if point $ D_{o} $ is selected anywhere within $ \triangle ABC $? Part 1: Construction of the points $ A_{1}, B_{1}, C_{1} $ Given a tetrahedron $ ABCD $ with vertex $ D $,see picture. Lines $d1$ and $d2$ are bisectors of $ \triangle ABC $. The crossing of $d1$ and $d2$ is the point $ D_{0} $.We draw the line $ D D_{0} $ and also the line $z$, parallel to $ D D_{0}$. In the plane $ (ABC) $, we draw the line $ p $, parallel to $ BC $. In this plane $ (ABC) $, the line $ p $ is crossing the 2 bisectors $d1$ and $d2$ in 2 points, where $b1$ and $b2$ are parallel to $ D D_{0}$. In the plane $ (b1, d1,D D_{0},C) $, the lines $ d1 $ and $ CD $ are crossing in the point $A1$. In the plane $ (b2, d2,D D_{0},B) $, the lines $ d2 $ and $ BD $ are crossing in the point $A2$. In the plane $ (p, b1, b2,z,A1, A2) $, the lines $ A1A2 $ and $ z $ are crossing in the point $A'$; plane $ (BCD) \cap z $ = $A'$. That point $A'$ is the asked point $ A_{1}$. The same construction can be used for the points $ B_{1} $ and $ C_{1} $. If the tetrahedron is regular, then the points $ A_{1}, B_{1}, C_{1} $ form a plane, parallel to the plane $ (ABC)$, see part 2. Part 2: Calculation of the coordinates of the points $ A_{1}, B_{1}, C_{1} $ Given a regular tetrahedron $ ABCD $ with vertex $ D $; each side has length 1. \[A(0,0,0), B(\frac{\sqrt{3}}{2},\frac{1}{2},0), C(0,1,0)\] \[D_{0}(\frac{\sqrt{3}}{6},\frac{1}{2},0), D(\frac{\sqrt{3}}{6},\frac{1}{2},\frac{\sqrt{6}}{3})\] In this regular tetrahedron: $ D D_{0} \bot (ABC)$, hence $ D D_{0} \bot (x\;y)$. The equation of a plane $(PQR)$ with \[P(x_{1},y_{1},z_{1}),Q(x_{2},y_{2},z_{2}),R((x_{3},y_{3},z_{3})\] is given by: \[ \left| \begin{array}{ccc} x- x_{1}& y- y_{1}& z- z_{1}\\ x_{2}- x_{1}& y_{2}- y_{1}& z_{2}- z_{1} \\ x_{3}- x_{1}& y_{3}- y_{1}& z_{3}- z_{1} \end{array} \right|=0\] a) equation of the plane $(DBC)$: \[ \left | \begin{array}{ccc} x& y- 1& z\\ \frac{\sqrt{3}}{2}& - \frac{1}{2}& 0 \\ \frac{\sqrt{3}}{6}& - \frac{1}{2}& \frac{\sqrt{6}}{3} \end{array} \right|=0\] \[\frac{\sqrt{6}}{6}x+\frac{\sqrt{2}}{2}(y-1)+\frac{\sqrt{3}}{6}z=0\] Through point $A$, we are drawing a parallel with $D D_{0}$, i.e. this parallel $\bot (x\;y)$ The equation of this parallel is: \[ \left\{\begin{array}{ll} x=0\\ y=0\end{array}\right.\] This parallel through $A$ $ \cap (DBC) = A'$: \[ \left\{\begin{array}{lll} \frac{\sqrt{6}}{6}x+\frac{\sqrt{2}}{2}(y-1)+\frac{\sqrt{3}}{6}z=0\\x=0\\ y=0\end{array}\right.\] \[A'(0,0,\sqrt{6})=A_{1}(0,0,\sqrt{6})\] b) equation of the plane $(ADB)$: \[ \left| \begin{array}{ccc} x& y& z\\ \frac{\sqrt{3}}{2}& \frac{1}{2}& 0 \\ \frac{\sqrt{3}}{6}& \frac{1}{2}& \frac{\sqrt{6}}{3} \end{array} \right|=0\] \[\frac{\sqrt{6}}{6}x-\frac{\sqrt{2}}{2}y+\frac{\sqrt{3}}{6}z=0\] Through point $C$, we are drawing a parallel with $D D_{0}$, i.e. this parallel $\bot (x\;y)$ The equation of this parallel is: \[ \left\{\begin{array}{ll} x=0\\ y=1\end{array}\right.\] This parallel through $C$ $ \cap (ADB) = C'$: \[ \left\{\begin{array}{lll} \frac{\sqrt{6}}{6}x-\frac{\sqrt{2}}{2}y+\frac{\sqrt{3}}{6}z=0\\x=0\\ y=1\end{array}\right.\] \[C'(0,1,\sqrt{6})\] c) equation of the plane $(ADC)$: \[ \left| \begin{array}{ccc} x& y& z\\ 0& 1& 0\\ \frac{\sqrt{3}}{6}& \frac{1}{2}& \frac{\sqrt{6}}{3} \end{array} \right|=0\] \[\frac{\sqrt{6}}{3}x-\frac{\sqrt{3}}{6}z=0\] Through point $B$, we are drawing a parallel with $D D_{0}$, i.e. this parallel $\bot (x\;y)$ The equation of this parallel is: \[ \left\{\begin{array}{ll} x=\frac{\sqrt{3}}{2}\\ y=\frac{1}{2}\end{array}\right.\] This parallel through $B$ $ \cap (ADC) = B'$: \[ \left\{\begin{array}{lll} \frac{\sqrt{6}}{3}x-\frac{\sqrt{3}}{6}z=0\\x=\frac{\sqrt{3}}{2}\\ y=\frac{1}{2}\end{array}\right.\] \[B'(\frac{\sqrt{3}}{2},\frac{1}{2},\sqrt{6})\] Because the value of the z-coordinate of $A',B'$ and $C'$ is $\sqrt{6}$, are the surface of $\triangle ABC$ and the surface of $\triangle A'B'C'$ equal. Because the height of the tetrahedron $ ABCD $ equals $\frac{\sqrt{6}}{3}$ (the z-coordinate of the point $D$) and the height of the tetrahedron $ A'B'C'D_{0} $ equals $\sqrt{6}$, is the volume of $ ABCD $ indeed one third the volume of $ A_{1}B_{1}C_{1}D_{0} $. Part 3: Calculation of another tetrahedron a) Given a arbitrary tetrahedron $ ABCD $ with vertex $ D $. \[A(0,0,0), B(a,b,0), C(0,c,0)\] Point $D_{0}$ is the crossing point of the bisectors of $\triangle ABC$. $D D_{0}$ is perpendicular on the plane $(ABC)$, points $D$ and $ D_{0}$ have the same x-coordinate and y-coordinate; the z-coordinate of $D$ is 1. The same calculation as in Part 2 is made with result: the volume of $ ABCD $ is one third the volume of $ A_{1}B_{1}C_{1}D_{0} $. b) Given a regular tetrahedron $ ABCD $ with vertex $ D $; each side has length 1. \[A(0,0,0), B(\frac{\sqrt{3}}{2},\frac{1}{2},0), C(0,1,0), D(\frac{\sqrt{3}}{6},\frac{1}{2},\frac{\sqrt{6}}{3})\] The point $ D_{o} $ is selected anywhere within $ \triangle ABC $: \[D_{0}(k,l,0)\] with $k \neq \frac{\sqrt{3}}{6}$ \ and/or \ $l \neq \frac{1}{2}$. The same calculation as in Part 2 is made with result: the volume of $ ABCD $ is NEVER one third the volume of $ A_{1}B_{1}C_{1}D_{0} $.

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