Monday, May 19, 2008

Geometry Problem 108



See complete Problem 108
Triangle, Angles, Median, Congruence, Cyclic Quadrilateral. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

5 comments:

  1. Think about a cyclic quadrilateral.

    Geometry problem solving is one of the most challenging skills for students to learn. When a problem requires auxiliary construction, the difficulty of the problem increases drastically, perhaps because deciding which construction to make is an ill-structured problem. By “construction,” we mean adding geometric figures (points, lines, planes) to a problem figure that wasn’t mentioned as "given."

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  2. Draw a perpendicular through D, wich intersects BC in E. Then triangle DEC is congruent to triangle DEA. That implies that DEA=90-alpha, wich shows that ABED is cyclic, so EBD=EAD. this is alpha=x.


    srry for my english :$

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  3. This problem x has two solutions
    x=Alpha
    x=90-Alpha

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  4. http://geometri-problemleri.blogspot.com/2009/06/problem-29-ve-cozumu.html

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  5. Let the perpendicular bisector of AC meet AB extended at E. Then < EAC = 90-x and so < AED = x = < DEC = < DBC

    Hence BDCE is cyclic and ABC is a right triangle with D the centre and so x = @

    Sumith Peiris
    Moratuwa
    Sri Lanka

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