document.write( "Question 1199525: Use coordinate geometry to prove the following statement. Given: △ABC; A(c, d), B(c, e), C(f, e) Prove: The circumcenter of △ABC is a point on the triangle. \n" ); document.write( "

Algebra.Com's Answer #833455 by greenestamps(13203) $\"\"$ $\"About$

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\n" ); document.write( "The response from the other tutor shows a basic standard geometric proof; it does not use coordinate geometry.

\n" ); document.write( "Coordinate geometry is a tool with which geometric proofs are sometimes far easier than with Euclidean geometry.

\n" ); document.write( "The circumcenter of any triangle lies on the intersection of the perpendicular bisectors of any two sides of the triangle.

\n" ); document.write( "Side AB of the given triangle is a vertical segment with midpoint (c,(d+e)/2), so the equation of the perpendicular bisector of AB is the horizontal line y=(d+e)/2.

\n" ); document.write( "Side BC of the given triangle is a horizontal segment with midpoint ((c+f)/2,e), so the equation of the perpendicular bisector of AB is the vertical line x=(c+f)/2.

\n" ); document.write( "The circumcenter of the triangle, the intersection of those two perpendicular bisectors, is the point ((c+f)/2,(d+e)/2).

\n" ); document.write( "But the point ((c+f)/2,(d+e)/2) is the midpoint of side AC of the triangle.

\n" ); document.write( "So the circumcenter of the triangle is a point on the triangle.

\n" ); document.write( "QED

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