Friday, December 14, 2012

Problem 832: Quadrilateral, Isosceles triangle, 70-70-40 degrees, Congruence, Auxiliary lines, Equilateral triangle

Geometry Problem
Problem submitted by Charles T.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 832.

Online Geometry Problem 832: Quadrilateral, Isosceles triangle, 70-70-40 degrees, Congruence, Auxiliary lines, Equilateral triangle

5 comments:

  1. http://img51.imageshack.us/img51/5048/problem832.png
    From D draw DE//BC and DE=BC=AD
    BCDE is a parallelogram => BE //DC => BE is angle bisector of angle ABD
    So ED=AD=AE => ADE is a quadrilateral triangle and ∠ ADE=60
    So ∠BDC=∠BDE=70-60=10

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    Replies
    1. Anonymous

      Within angle ABD build angle equilateral APD , BP perpendicular bisector for the segment AD .
      BP parallel with DC , DP=BC than BPDC trapezoid or parallelogram,
      BPD=150grade when ABC <90 we have parallelogram, and angle DPC is 10 grade .
      When 90<ABC<180 it becomes that DPC=130 grade

      Erina New Jersey

      Delete
  2. Complete the rectangle ADCE. Join BE.
    Triangle BEC is equilateral.
    Since triangles BEA,BCD are congruent,
    angles ABE,DBC are equal each equal to
    (60 - 40)/2 = 10 degrees.

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  3. Prof Radu Ion,Sc.Gim.Bozioru,BuzauDecember 17, 2012 at 12:26 PM

    Construim paralelogramul ADMB si vom triunghiul echilateral BMC deoarece triunghiurile ABC=BDM si BCD=MCD=>m(<CBD)=m(<MBD)-m(<CBD)=70-60=10

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  4. On the angle bisector of B mark E such that BE = CD. But BE is also parallel to CD hence BECD is a parelleligram and DE = and is oarellel to BC = AE SAS
    Hence Tr. AED is equilateral and x = < EDB = 10

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete