Thursday, September 22, 2011

Problem 671: Triangle, Cevian, Angles, Congruence

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

Click the figure below to see the complete problem 671.

Online Geometry Problem 671: Triangle, Cevian, Angles, Congruence

4 comments:

  1. http://img838.imageshack.us/img838/3824/problem671.png
    Draw triangle BDE as image of triangle BDA ( see picture)
    Note that BE=CD and (DCB)=(EBC)=4. alpha
    And DBCE is an isosceles trapezoid
    (BDE)=(BDA)= 180-7.alpha
    In triangle ABD , x+180-7.alpha+3.alpha= 180 >> x=4. alpha
    In triangle ABC , 7.alpha+4.alpha+4.alpha=180 >> alpha=10 and x=40
    Peter Tran

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  2. Okay, I have a really simple solution.

    At first the problem seems to be awful but it can be taken down faster.

    Let P be a point on BC so that DB = DP, hence <CDP = 3a and then ∆CDP = ∆ABD and x = 4a, hence a = 10° whereat x = 40°.

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  3. Find E on DC such that < CBE = 3@
    Let BD and BE extended meet the circle ABC at X and Y.

    AX = CY so ACYX is an isoceles trapezoid.

    Now < AEB = 7@ = < ABE

    Hence AB = c = AE = CD
    So AD = CE = b-c
    Further < XAC = YCA
    So tr.s AXD and CYE are congruent SAS.

    < BDC = < BEA = 7@

    Hence 7@ + 7@ + 4@ = 180 and therefore

    @ = 10 and x = 40

    Sumith Peiris
    Moratuwa
    Sri Lanka

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