Difference between revisions of "Steiner's Theorem"
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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 | + | 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 03:02, 23 March 2013
Steiner's Theorem states that in a trapezoid with
and
, we have that the midpoint of
and
, the intersection of diagonals
and
, and the intersection of the sides
and
are collinear.
Proof
Let be the intersection of
and
,
be the midpiont of
,
be the midpoint of
, and
be the intersection of
and
. We now claim that
. First note that, since
and
[this is because
], we have that
. Then
, and
, so
. We earlier stated that
, so we have that
from SAS similarity. We have that
,
, and
are collinear, and since
and
are on the same side of line
, we can see that
from SAS. Therefore
, so
,
, and
are collinear.
Now consider triangles and
. Segments
and
are transversal lines, so it's not hard to see that
. It's also not hard to show that
by SAS similarity. Therefore
, which implies that
,
, and
are collinear. This completes the proof.
Alternative Proof
Define the same as above. Then there exists a positive homothety about
mapping
the homothety takes
are collinear. Similarly, there is a negative homothety taking
are collinear. So,
are collinear, as desired
.
Proof by IDMasterz