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  1. #1
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    geometry

    I was wondering if someone could show me how to do prove these in a Euclidean way:

    I need to show that the following three statements are equivalent for the triangle ABC,
    gamma (the angle opposite AB) = 90 degrees
    C is the orthocenter of ABC
    the midpoint of AB is the circumcenter of ABC

    I think I can prove this if I show the first implies the second, the second implies the third, and the third implies the first. I drew out diagrams and I see that it makes sense, but I'm having trouble writing it out in a more rigorous way.
    Any help would be appreciated. Thanks in advance.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by PvtBillPilgrim View Post
    I was wondering if someone could show me how to do prove these in a Euclidean way:

    I need to show that the following three statements are equivalent for the triangle ABC,
    (1) gamma (the angle opposite AB) = 90 degrees
    (2) C is the orthocenter of ABC
    (3) the midpoint of AB is the circumcenter of ABC

    I think I can prove this if I show the first implies the second, the second implies the third, and the third implies the first. I drew out diagrams and I see that it makes sense, but I'm having trouble writing it out in a more rigorous way.
    Any help would be appreciated. Thanks in advance.
    (1) => (2) since \angle C=\gamma=90^0, then AC (BC) is the altitude w/ respect to vertex A (B) [which both contain C]. The altitude with respect to C passes also C. The three altitudes intersect at C, therefore, C is the orthocenter of \Delta ABC.

    (2) => (3) i'll try to work on this tomorrow.

    (3) => (1) Let D, the midpoint of AB, be the circumcenter of ABC. Let E be on CB such that DE is the perpendicular bisector of CB, and let F be on CA such that DF is the perpendicular bisector of CA. Then, CFDE forms a quadrilateral. In particular, since angles CDF and DEC forms right angles, then CFDE is a rectangle. therefore, angle C is a right angle.
    Last edited by kalagota; July 17th 2008 at 12:59 AM.
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  3. #3
    MHF Contributor kalagota's Avatar
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    (2) => (3) Let C be the orthocenter of ABC. then the altitude from A to BC is AC which means AC forms a right angle with BC. Let M_{AB} be the midpoint of AB and consider a point outside the triangle, say D, such that ADBC forms a rectangle. Then, AB is a diagonal of the rectangle.
    Note that the diagonals of a rectangle bisect each other. Thus, CM_{AB} is in the diagonal CD. Therefore CM_{AB} is congruent to M_{AB}B.
    Let M_{CB} be on CB such that M_{AB}M_{CB} perpendicular to CB.

    now, M_{AB}M_{CB}=M_{AB}M_{CB}; and CM_{AB} is congruent to M_{AB}B. Therefore, by HL theorem, \Delta CM_{CB}M_{AB} \cong \Delta BM_{CB}M_{AB} (noting that the triangles are right.). By CPCTC Theorem, CM_{CB} \cong BM_{CB}; therefore, M_{AB}M_{CB} is the perpendicular bisector of CB.

    similarly, if M_{CA} be on CA such that M_{AB}M_{CA} perpendicular to CA, it can be shown that M_{AB}M_{CA} is the perpendicular bisector of CA.

    The perpendicular bisector of AB passes M_{AB}. Since all the perpendicular bisectors of each side intersect at M_{AB}, therefore M_{AB} is the orthocenter of ABC..


    PS: drawing will help you in the visualization of the proof..
    Last edited by kalagota; July 17th 2008 at 01:50 AM.
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  4. #4
    MHF Contributor kalagota's Avatar
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    this is the drawing..
    Attached Thumbnails Attached Thumbnails geometry-rep.gif  
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