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

Posted by Antonio Gutierrez at 3:57 PM
Labels: , , , , ,

9 comments:

  1. Peter TranDecember 14, 2012 at 5:31 PM

    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

    ReplyDelete
    Replies
    1. AnonymousDecember 30, 2012 at 9:34 PM

      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. PravinDecember 15, 2012 at 8:26 AM

    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.

    ReplyDelete
  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

    ReplyDelete
  4. Sumith PeirisJuly 15, 2015 at 4:07 AM

    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
  5. Sumith PeirisJune 9, 2017 at 2:40 AM

    Solution 2

    Complete equilateral triangle AXD, X inside ABD

    Triangles ABX and CBX are congruent SSS. So < DBX = 20 = < BDC

    Hence XD = BC and XD // BC and so BCDX is a //ogram

    Therefore x = < XDB = 10

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. Sumith PeirisJune 9, 2017 at 2:40 AM

    This comment has been removed by the author.

    ReplyDelete
  7. Geek37November 13, 2018 at 3:15 AM

    Here is my solution...

    https://www.youtube.com/watch?v=274_UUFOcLo

    ReplyDelete
  8. MarcoFebruary 10, 2024 at 9:14 PM

    Let <CBD=x, <BCD=160-x
    sin(160-x)/BD=sin20/BC
    BC/BD=sin20/sin(160-x)
    <ABD=40
    sin40/AD=sin70/BD
    AD/BD=sin40/sin70
    AD=BC
    sin20/sin(160-x)=sin40/sin70
    sin20sin70=sin(160-x)sin40
    sin20cos20=sin(160-x)sin40
    sin(160-x)=1/2
    160-x=30 or 150
    x=130 (rej.) or 10

    ReplyDelete

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