Prove that, if in a tetrahedron $ABCD$ the heights are congruent( of equal lengths) and $A$ is projected on the $(BCD)$ plane in the orthocenter of triangle $BCD$, then $ABCD$ is a regular tetrahedron.
I have noticed from this answer, Concurrency of the heights of a tetrahedron with opposite edges perpendicular., that that by projecting $A$ in the orthocenter, we also could argue that the heights are concurent, and then the base of the tetrahedron is equilateral. However, this problem just eludes me, and nothing I searched in textbooks helps. Any help, please?
Best Answer
Let ABCD is tetrahedron with congruent heights. Volume of tetrahedron is equal $\frac{1}{3}h_i S_i$, where $h_i$ and $S_i$ are corresponding heights and faces areas. As $h_i$ are equal, so $S_i$ are also equal.
Let $A_1$ is projection of $A$ on the $(BCD)$ plane, and $A_1$ is the orthocenter of triangle $BCD$.
Consider altitude $BE$ of triangle $BCD$. $$AA_1 \perp (BCD), A_1E\perp CD \Rightarrow AE \perp CD$$ Then $AE$ is altitude of triangle $ACD$. Area of faces $BCD$ and $ACD$ are equal $CD\cdot BE / 2$ and $CD \cdot AE / 2$ and are equal to each other. Then $AE=BE$. Then triangles $CEB$ and $CEA$ are congruent and triangles $DEB$ and $DEA$ are congruent. Then $AC=BC$ and $AD=BD$.
Using the same method with altitude $CF$ passing through $A_1$, one can obtain $AB=BC$, $AD=CD$. Using the same method with altitude $DG$ passing through $A_1$, one can obtain $AC=CD$, $AB=BD$. Then $AB=AC=BC=CD=AD=BD$. Then tetrahedron ABCD is regular.