# Steiner's Theorem

Steiner's Theorem states that in a trapezoid $ABCD$ with $AB\parallel CD$ and $AD\nparallel BC$, we have that the midpoint of $AB$ and $CD$, the intersection of diagonals $AC$ and $BD$, and the intersection of the sides $AD$ and $BC$ are collinear.

## Proof

Let $E$ be the intersection of $AD$ and $BC$, $F$ be the midpiont of $AB$, $G$ be the midpoint of $CD$, and $H$ be the intersection of $AC$ and $BD$. We now claim that $\triangle EAF \sim \triangle EDG$. First note that, since $\angle DEC=\angle AEB$ and $\angle EAB=\angle EDC$ [this is because $AB\parallel CD$], we have that $\triangle EDC\sim \triangle EAB$. Then $\frac{EA}{AB}=\frac{ED}{DC}$, and $\frac{AF}{AB}=\frac{DG}{DC}=\frac{1}{2}$, so $\frac{EA}{AF}=\frac{ED}{DG}$. We earlier stated that $\angle EDC=\angle EAB$, so we have that $\triangle EAF \sim \triangle EDG$ from SAS similarity. We have that $E$, $A$, and $D$ are collinear, and since $F$ and $G$ are on the same side of line $AD$, we can see that $\triangle EAF \sim \triangle EDG$ from SAS. Therefore $EF \parallel EG$, so $E$, $F$, and $G$ are collinear.

Now consider triangles $HBA$ and $HDC$. Segments $AC$ and $BD$ are transversal lines, so it's not hard to see that $\triangle HAB \sim \triangle HCD$. It's also not hard to show that $\triangle HBE \sim \triangle HDF$ by SAS similarity. Therefore $\angle BHE=\angle DHF$, which implies that $F$, $G$, and $H$ are collinear. This completes the proof.

Alternative Proof

Define $E, F, G, H$ the same as above. Then there exists a positive homothety about $E$ mapping $\triangle EAB \to \triangle ECD \implies$ the homothety takes $F \to G \implies E, F, G$ are collinear. Similarly, there is a negative homothety taking $\triangle HAB \to \triangle HCD \implies H, F, G$ are collinear. So, $E, F, G, H$ are collinear, as desired $\blacksquare$.

Proof by IDMasterz