Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Steiner's Theorem"

(Got kicked out of computer lab because of halo players; came back to finish the proof)
m (→‎Proof)
 
(2 intermediate revisions by 2 users not shown)
Line 3: Line 3:
 
Let <math>E</math> be the intersection of <math>AD</math> and <math>BC</math>, <math>F</math> be the midpiont of <math>AB</math>, <math>G</math> be the midpoint of <math>CD</math>, and <math>H</math> be the intersection of <math>AC</math> and <math>BD</math>. We now claim that <math>\triangle EAF \sim \triangle EDG</math>. First note that, since <math>\angle DEC=\angle AEB</math> and <math>\angle EAB=\angle EDC</math> [this is because <math>AB\parallel CD</math>], we have that <math>\triangle EDC\sim \triangle EAB</math>. Then <math>\frac{EA}{AB}=\frac{ED}{DC}</math>, and <math>\frac{AF}{AB}=\frac{DG}{DC}=\frac{1}{2}</math>, so <math>\frac{EA}{AF}=\frac{ED}{DG}</math>. We earlier stated that <math>\angle EDC=\angle EAB</math>, so we have that <math>\triangle EAF \sim \triangle EDG</math> from SAS similarity. We have that <math>E</math>, <math>A</math>, and <math>D</math> are collinear, and since <math>F</math> and <math>G</math> are on the same side of line <math>AD</math>, we can see that <math>\triangle EAF \sim \triangle EDG</math> from SAS. Therefore <math>EF \parallel EG</math>, so <math>E</math>, <math>F</math>, and <math>G</math> are collinear.
 
Let <math>E</math> be the intersection of <math>AD</math> and <math>BC</math>, <math>F</math> be the midpiont of <math>AB</math>, <math>G</math> be the midpoint of <math>CD</math>, and <math>H</math> be the intersection of <math>AC</math> and <math>BD</math>. We now claim that <math>\triangle EAF \sim \triangle EDG</math>. First note that, since <math>\angle DEC=\angle AEB</math> and <math>\angle EAB=\angle EDC</math> [this is because <math>AB\parallel CD</math>], we have that <math>\triangle EDC\sim \triangle EAB</math>. Then <math>\frac{EA}{AB}=\frac{ED}{DC}</math>, and <math>\frac{AF}{AB}=\frac{DG}{DC}=\frac{1}{2}</math>, so <math>\frac{EA}{AF}=\frac{ED}{DG}</math>. We earlier stated that <math>\angle EDC=\angle EAB</math>, so we have that <math>\triangle EAF \sim \triangle EDG</math> from SAS similarity. We have that <math>E</math>, <math>A</math>, and <math>D</math> are collinear, and since <math>F</math> and <math>G</math> are on the same side of line <math>AD</math>, we can see that <math>\triangle EAF \sim \triangle EDG</math> from SAS. Therefore <math>EF \parallel EG</math>, so <math>E</math>, <math>F</math>, and <math>G</math> are collinear.
  
−
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</math> \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$ 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.
  
 +
'''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 H, F, G</math> are collinear. So, <math>E, F, G, H</math> are collinear, as desired <math>\blacksquare</math>.
 +
 +
''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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Steiner%27s_Theorem&oldid=51892"