Proof question help. (1 Viewer)

zacn

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Can anyone help answer this question.

Prove by contradiction that given three non-collinear points in a plane, there exists one and only one circle through these three points.
 

Pedro123

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Here's a brief rundown, reply if you want a more detailed response.
Take 3 points A, B and C. Draw lines AB and BC, labelling their midpoints z and y respectively. Then, draw perpendicular lines at Z and Y such that they intersect with each other at a point (say O).
You can easily prove congruence of OZB and OZA, meaning OB = OA. Repeat for OYB and OYA, and you get that OA = OB = OC, meaning that the circle has centre O. It's trivial that there is only 1 circle through this point that satisfies this. This is the only one circle.
Now, assume there is another midpoint N. Drop perpendiculars from N to AB and BC to T and S. Using the fact that NA = NB = NC, reverse the steps above to get that T = Z and S = Y, which means that N = O, which is contradictory to the fact that N and O should be distinct.
 

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