Difference between revisions of "Steiner's Theorem"

Latest revision as of 03:02, 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

@@ Line 7: / Line 7: @@
 '''Alternative Proof'''
-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 HFG</math> are collinear. So, <math>E, F, G, H</math> are collinear, as desired <math>\blacksquare</math>.
+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>.
+''Proof by IDMasterz''
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Difference between revisions of "Steiner's Theorem"

Latest revision as of 03:02, 23 March 2013

Proof

See also