Saturday, October 17, 2009

Problem 367. Square, Angle, 45 degrees, Pythagoras, Congruence, Similarity

Proposed Problem
Click the figure below to see the complete problem 367 about Square, Angle, 45 degrees, Pythagoras, Congruence, Similarity.

Problem 367. Square, Angle, 45 degrees, Pythagoras, Congruence, Similarity.
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Complete Geometry Problem 367
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 9:26 AM
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3 comments:

  1. AB=BC=CD=AD=(a+b+x)/sqrt(2)...(*)

    from similarity between triangles AGD and HGA:
    AG=sqrt(x^2+xb)...(**)

    from similarity between triangles ABH and GAH:

    AB/GA=AH/GH=BH/AH

    by setting the values from (*) and (**); we get:
    AH=x*[(a+b+x)/sqrt(2)]/sqrt(x^2+bx)...(1).
    and
    AH=sqrt(x^2+xa)...(2).

    after solving the equations (1),(2) we get:

    x^2=a^2+b^2

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  2. http://geometri-problemleri.blogspot.com/2009/11/problem-47-ve-cozumu.html

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  3. since angle(GAF)=angle(GDF)=45 and angle(ADF)=90, GADF is a cyclic and so EGHF is an also cyclic. Hence two triangles BEG and BHC are similar and hence a(a+x)=BE(a+b+x)/sqrt(2).
    Since BE=(a+b+x)/{sqrt(2)(x+b)}, we have
    get x^2 =a^2 +b^2.

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