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Difference between revisions of "Steiner's Theorem"

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Now consider triangles <math>HBA</math> and <math>HDC</math>. Segments <math>AC</math> and <math>BD</math> are transversal lines, so it's not hard to see that <math>\triangle HAB \sim \triangle HCD</math>. It's also not hard to show that <math>\triangle HBE \sim \triangle HDF</math> by SAS similarity. Therefore <math>\angle BHE=\angle DHF</math>, which implies that <math>F</math>, <math>G</math>, and <math>H</math> are collinear. This completes the proof.
 
Now consider triangles <math>HBA</math> and <math>HDC</math>. Segments <math>AC</math> and <math>BD</math> are transversal lines, so it's not hard to see that <math>\triangle HAB \sim \triangle HCD</math>. It's also not hard to show that <math>\triangle HBE \sim \triangle HDF</math> by SAS similarity. Therefore <math>\angle BHE=\angle DHF</math>, which implies that <math>F</math>, <math>G</math>, and <math>H</math> are collinear. This completes the proof.
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'''Alternative Proof'''
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Define <math>E, F, G, H</math> the same as above. Then there exists a positive homothety about <math>E</math> mapping <math>\triangle EAB \to \triangle ECD \implies</math> the homothety takes <math>F \to G \implies E, F, G</math> are collinear. Similarly, there is a negative homothety taking <math>\triangle HAB \to \triangle HCD \implies H, F, G</math> are collinear. So, <math>E, F, G, H</math> are collinear, as desired <math>\blacksquare</math>.
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''Proof by IDMasterz''
  
 
==See also==
 
==See also==

Latest revision as of 02:02, 23 March 2013

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

See also

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