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  • #24696
    Pinaki Biswas
    Participant

    Show that in any triangle, a circumcircle exists.

    #24709
    swastik pramanik
    Participant

    This is just proving that:

    Theorem:

    Given any three non-collinear points \(A, B, C\) there exists a unique circle passing through \(A, B, C\).

    Proof:

    Let us consider three vertices of \(\Delta ABC\) i.e. \(A, B, C\). Suppose the perpendicular bisectors of \(BC\) and \(CA\) meet at \(S\). Then \(S\) lies on the perpendicular bisector of \(BC\) implies that \(SB=SC\). Again, \(S\) is also a perpendicular bisector of \(CA\) which implies that \(SC=SA\). Hence, we have \(SA=SB=SC\). Further more, any point equidistant from \(A, B, C\) should lie on the perpendicular bisectors of \(BC, CA\). Therefore, \(S\) is the only point equidistant from \(A, B, C\) and so the circle with centre \(S\) and radius \(SA\) is the unique circle passing through \(A, B, C\).

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