Difference between revisions of "Circumcenter"
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[[Image:Circumcircle2.PNG|center]] | [[Image:Circumcircle2.PNG|center]] | ||
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| + | == Proof that the perpendicular bisectors are concurrent == | ||
| + | We start with a diagram: | ||
| + | [[Image:Circumproof1.PNG|center]] | ||
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| + | One of the most common techniques for proving the [[concurrency]] of [[line]]s 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 <math>DE, EF</math> and <math>FD</math> (i.e. the [[medial triangle]] of <math>ABC</math>) does the trick. | ||
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| + | [[Image:Circumproof2.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). | ||
Revision as of 00:50, 18 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 are concurrent
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).
