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.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52814)   (Show Source):
You can put this solution on YOUR website! .
The side AB of the triangle is vertical x = c.
The side BC of the triangle is horizontal y = e.
So, the triangle ABC is a right-angled triangle.
Its right angle is at vertex B; its hypotenuse is AC (opposite to the right angle).
In any right-angled triangle, the center of the circumscribed circle is the mid-point of the hypotenuse.
It proves the statement.
Solved.
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
The response from the other tutor shows a basic standard geometric proof; it does not use coordinate geometry.
Coordinate geometry is a tool with which geometric proofs are sometimes far easier than with Euclidean geometry.
The circumcenter of any triangle lies on the intersection of the perpendicular bisectors of any two sides of the triangle.
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.
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.
The circumcenter of the triangle, the intersection of those two perpendicular bisectors, is the point ((c+f)/2,(d+e)/2).
But the point ((c+f)/2,(d+e)/2) is the midpoint of side AC of the triangle.
So the circumcenter of the triangle is a point on the triangle.
QED
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