Difference between revisions of "Circumcenter"
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[[Image:Circumcircle2.PNG|center]] | [[Image:Circumcircle2.PNG|center]] | ||
| − | == Proof that the perpendicular bisectors are concurrent == | + | == Proof that the perpendicular bisectors of a triangle are concurrent == |
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| + | === First Proof === | ||
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| + | We consider a [[triangle]] <math>\displaystyle ABC</math>. Let <math>\displaystyle O</math> be the intersection of the perpendicular bisectors of the [[side]]s <math>\displaystyle AB</math> and <math>\displaystyle BC</math>. Since <math>\displaystyle O</math> lies on the perpendicular bisector of <math>\displaystyle AB</math>, it is [[equidistant]] from <math>\displaystyle A</math> and <math>\displaystyle B</math>; likewise, it is equidistant from <math>\displaystyle B</math> and <math>\displaystyle {C}</math>. Hence <math>\displaystyle O</math> is equidistant from <math>\displaystyle A</math> and <math>\displaystyle {C}</math>; hence <math>\displaystyle O</math> also lies on the perpendicular bisector of <math>\displaystyle AC</math> (and is the circumcenter). | ||
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| + | === Second Proof === | ||
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We start with a diagram: | We start with a diagram: | ||
[[Image:Circumproof1.PNG|center]] | [[Image:Circumproof1.PNG|center]] | ||
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By [[SAS Similarity]] <math>\triangle BFD\cong \triangle BAC</math>. Thus <math>\angle BFD = \angle BAC</math> making <math>FD || AC</math>. Since <math>EO\perp AC</math> and <math>AC\| FD, EO\perp FD</math> making <math>EH</math> an [[altitude]] of <math>DEF</math>. Likewise, <math>DG</math> and <math>FI</math> are also altitudes. Thus, the problem is reduced to proving that the altitudes of a triangle are concurrent. This can be done using Ceva's Theorem (see [[orthocenter]] for more details). | By [[SAS Similarity]] <math>\triangle BFD\cong \triangle BAC</math>. Thus <math>\angle BFD = \angle BAC</math> making <math>FD || AC</math>. Since <math>EO\perp AC</math> and <math>AC\| FD, EO\perp FD</math> making <math>EH</math> an [[altitude]] of <math>DEF</math>. Likewise, <math>DG</math> and <math>FI</math> are also altitudes. Thus, the problem is reduced to proving that the altitudes of a triangle are concurrent. This can be done using Ceva's Theorem (see [[orthocenter]] for more details). | ||
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| + | It is worth noting that the existance of the circumcenter is a much more fundamentally important theorem than it might seem, since it implies that three [[point]]s determine a circle. | ||
Revision as of 19:21, 25 August 2006
The circumcenter is the center of the circumcircle of a polygon. However, it should be noted that only certain polygons can be circumscribed by a circle. All triangles have a circumcircle whose circumcenter is the intersection of the triangle's perpendicular bisectors. Quadrilaterals which have circumcircles are called cyclic quadrilaterals. Also, every regular polygon is cyclic.
Proof that the perpendicular bisectors of a triangle are concurrent
First Proof
We consider a triangle 
. Let 
be the intersection of the perpendicular bisectors of the sides 
and 
. Since lies on the perpendicular bisector of , it is equidistant from 
and 
; likewise, it is equidistant from and 
. Hence is equidistant from and ; hence also lies on the perpendicular bisector of 
(and is the circumcenter).
Second Proof
We start with a diagram:
One of the most common techniques for proving the concurrency of lines is Ceva's Theorem. However, there aren't any cevians in the diagram which would be needed for a direct application of Ceva's Theorem. Thus, we look for a way to make some by drawing in helpful lines. Drawing in 
and 
(i.e. the medial triangle of 
) does the trick.
By SAS Similarity 
. Thus 
making 
. Since 
and 
making 
an altitude of 
. Likewise, 
and 
are also altitudes. Thus, the problem is reduced to proving that the altitudes of a triangle are concurrent. This can be done using Ceva's Theorem (see orthocenter for more details).
It is worth noting that the existance of the circumcenter is a much more fundamentally important theorem than it might seem, since it implies that three points determine a circle.
