Friday, June 29, 2012

Problem 775: Triangle, Isogonal lines, Congruent angles, Perpendicular, Metric Relations

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 775.

Online Geometry Problem 775: Triangle, Isogonal lines, Congruent angles, Perpendicular, Metric Relations.

4 comments:

  1. Angle EBM = Angle DBF
    So Rt triangles EMB and DFB are similar
    ∴ x : 8 = BE : BD
    Next angle HBE = angle GBD
    So Rt triangles HBE and GBD are similar
    ∴ BE : BD = EH : DG = 25 : 28
    Follows x : 8 = 25 : 28
    Hence x = 50/7

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  2. Triangle BEM and BFD are similar (AA)
    Triangle BDG and BHE are similar (AA)
    Hence DG/EH =BD/BE =ED/EM
    28/25 =8/EM
    So EM =50/7

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  3. Triangle BDG similar to ∆ BEH ….. ( Case AA)
    So BD/BE= 28/25
    Triangle BFD similar to ∆ BME ….. ( case AA)
    So DF/EM=8/x=BD/BE=28/25
    So x=50/7

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  4. Let be P on BD such that BP=BE, and let be Q and N the perpendiculars from P to AB and BC respectively. Note that BPQ and BME are congruent, and that BPQ is similar with BDF and BNP is similar with BGB (AA). Hence x=50/7

    Greetings go-solvers.

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