I'm having trouble solving this problem, can someone help?
In tetrahedron $ABCD$, the sum of the areas of faces $ABC$ and $ABD$ is equal to the sum of the areas of faces $ACD$ and $BCD$. Let $E$, $F$, $G$, and $H$ be the midpoints of sides $BC$, $AC$, $AD$, and $BD$, respectively, and let $I$ be the incenter of tetrahedron $ABCD$. Prove that points $E$, $F$, $G$, $H$, and $I$ are coplanar.
I've correctly proven that $E$, $F$, $G$, and $H$ are coplanar and that $EFGH$ is a parallelogram, but I'm stuck on how to prove that $I$ is also on the plane of this parallelogram.
Best Answer
Work with barycentric coordinates, the coordinate of a point $P$ is $P=[(PBCD):(APCD):(ABPD):(ABCP)]$, the ratio of (signed) volumes of the sub-tetrahedra. Clearly \begin{align*} A&=[1:0:0:0]\\ B&=[0:1:0:0]\\ C&=[0:0:1:0]\\ D&=[0:0:0:1] \end{align*} and we have, since they are mid-points, \begin{align*} E&=[0:\tfrac12:\tfrac12:0]\\ F&=[\tfrac12:0:\tfrac12:0]\\ G&=[\tfrac12:0:0:\tfrac12]\\ H&=[0:\tfrac12:0:\tfrac12] \end{align*} We see EFGH is the plane such that the sum of the weights of the first two coordinates is the sum of weights of the last two. But $$ I=[\Delta_{BCD}:\Delta_{ACD}:\Delta_{ABD}:\Delta_{ABC}] $$ (since volume of $PBCD$ is the area of the base $BCD$ times one-third of the height (which is the inradius so this is just a common constant of proportionality) and we are given the sum of areas $\Delta_{BCD}+\Delta_{ACD}=\Delta_{ABD}+\Delta_{ABC}$.