Friday, February 12, 2010

Geometry Problem 425: Quadrilateral, Triangle, Angles, 10, 20, 50 degrees, Congruence

Proposed Problem
Click the figure below to see the complete problem 425 about Quadrilateral, Triangle, Angles, 10, 20, 50 degrees, Congruence.

Problem 425: Quadrilateral, Triangle, Angles, 10, 20, 50 degrees, Congruence.
See also:
Complete Problem 425

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 10:39 AM
Labels: , , , , , ,

9 comments:

  1. AnonymousFebruary 12, 2010 at 3:16 PM

    http://geometri-problemleri.blogspot.com/2010/02/problem-73-ve-cozumu.html

    ReplyDelete
  2. AnonymousApril 11, 2010 at 10:18 AM

    Here is a diferent solution from the previous post:

    http://ricardosandoval.wordpress.com/2010/04/11/solucao-em-ingles-para-um-problema-da-internet/

    ReplyDelete
  3. AnonymousApril 11, 2010 at 11:40 AM

    Another solution that constructs only one new point:

    http://ricardosandoval.wordpress.com/2010/04/11/solucao-em-ingles-para-um-problema-da-internet/

    And doesn't uses circle properties as in:

    http://amenazatriangular.blogspot.com/

    ReplyDelete
  4. geometriMay 7, 2010 at 11:59 PM

    solution
    http://cebirci.net/forum/ozel-geometri-sorulari/10-20-50-x.html#277
    by hanzala

    ReplyDelete
  5. AnonymousAugust 18, 2014 at 10:09 AM

    Drawing circumcircle of BCD turns the problem into a cevian model 20,10,30,20,120-x,x-20.
    General solution of that kind of cevian model listed here: http://geomania.org/forum/fantezi-geometri/ucgen-icerisinde-p-noktasi-model-4-3/ but in Turkish. But you can follow via pictures.
    Remember that in a cevian model, some angles can be interchanged.

    ReplyDelete
  6. Sumith PeirisDecember 28, 2015 at 3:31 AM

    Draw perpendiculars BX from B to DC and DY from D to AB

    < DAC = 20 and < BDY = 40 = < BDC hence Tr.s BDY and BDX are congruent ASA

    Tr. ADY is 30-60-90 and so DY = DA /2 = DC/2 = DX. Hence DX = XC and so Tr. BDC is isoceles and x = 100

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. APOSTOLIS MANOLOUDISJune 18, 2016 at 8:36 AM

    Problem 425
    Dialing the BDE equilateral triangle, so < DEB = 60 = 2.30 = 2.<DAB, therefore AE = ED = DB = EB , <EAB = 10=<EBA.But triangle AED=triangle CDB( AD=DC,ED=DB,<ADE=40=<CDB ).So
    <BCD=<EAD=40,then <DBC=<DEA=100.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  8. AnonymousMay 2, 2020 at 8:03 PM

    The problem is true for all t, 60-t, 30-t.

    Let $BDC$ circumcircle meets AC at P. It is same cevian problem with:
    https://output.jsbin.com/fofecum#20,10,30,20

    ReplyDelete
  9. MarcoDecember 15, 2023 at 10:09 PM

    Let H be the intersecting point of BD & AC
    <BHA=120
    <BCH=120-x
    <BCD=140-x

    In triangle BCD
    sinx/CD=sin(140-x)/BD
    BD/CD=sin(140-x)/sinx---------(1)

    AD=CD
    <DAC=<DCA=20
    <BAD=30
    In triangle BAD
    sin30/BD=sin50/AD
    BD/AD=sin30/sin50------------(2)

    Since AD=CD, (1)=(2)
    sin(140-x)/sinx=sin30/sin50
    sin30sinx=sin50sin(140-x)
    sinx=2sin50sin(140-x)
    sinx=cos(90-x)-cos(190-x)
    cos(190-x)=0
    190-x=90
    x=100

    ReplyDelete

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