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


    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

  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°.