Let $M$ and $N$ are midpoint of edges $AB$ and $CD$ of the tetrahedron $ABCD$, $AN=DM$ and $CM=BN$. Prove that $AC=BD$. S. Berlov
Problem
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Tags: geometry, 3D geometry, tetrahedron, parallelogram, geometry proposed
Luis González
17.04.2014 09:27
Construct parallelograms $CMDY$ and $ANBZ$ $\Longrightarrow$ $DY=CM=BN=AZ,$ $MY=NZ=2 \cdot MN$ and since $AN=DM,$ then it follows that $\triangle DYM \cong \triangle AZN$ by SSS $\Longrightarrow$ $DN=AM=BM.$ Now construct parallelograms $CBDU$ and $ACBV$ $\Longrightarrow$ $DU=BC,$ $UN=BN=CM$ and since $DN=BM,$ it follows that $\triangle DUN \cong \triangle BCM$ by SSS $\Longrightarrow$ $BD=VB=AC.$
onyqz
05.10.2024 16:58
We will use Stewart's Theorem, first use it on $\triangle ACD$:
$$\frac{1}{4}CD^3+AN^2\cdot CD = AC^2\cdot \frac{1}{2}CD + AD^2\cdot \frac{1}{2}CD \, ,$$so $AN^2=AC^2+AD^2-\frac{1}{2}CD^2$.
Similarly on triangle $\triangle ADB$: $DM^2=AD^2+BD^2-\frac{1}{2}AB^2$. Since $AN^2=DM^2$ we have $$AC^2+\frac{1}{2}AB^2=BD^2+\frac{1}{2}CD^2$$or equivalently $$AC^2=BD^2+\frac{1}{2}CD^2-\frac{1}{2}AB^2$$.
For $\triangle ABC$: $CM^2=AC^2+BC^2-\frac{1}{2}AB^2$ and for $\triangle BCD$: $BN^2=BD^2+BC^2-\frac{1}{2}CD^2$. Therefore, since $CM^2=BN^2$ we also have $$AC^2+\frac{1}{2}CD^2=BD^2+\frac{1}{2}AB^2$$or equivalently
$$AC^2=BD^2+\frac{1}{2}AB^2-\frac{1}{2}CD^2$$.
Hence $BD^2+\frac{1}{2}CD^2-\frac{1}{2}AB^2=BD^2+\frac{1}{2}AB^2-\frac{1}{2}CD^2$ from which $AB=CD$. Thus $AC^2+\frac{1}{2}CD^2=BD^2+\frac{1}{2}AB=BD^2+\frac{1}{2}CD^2$, so $AC=BD$.
This concludes the problem. $\square$