Thursday, January 21, 2010

Problem 421: Right Triangle, Cevian, Angles, Measurement

Proposed Problem
Click the figure below to see the complete problem 421 about Right Triangle, Cevian, Angles, Measurement.

Problem 421: Right Triangle, Cevian, Angles, Measurement.
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Complete Problem 421
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 6:24 PM
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12 comments:

  1. AnonymousJanuary 21, 2010 at 8:41 PM

    http://geometri-problemleri.blogspot.com/2010/01/problem-63-ve-cozumu.html

    ReplyDelete
  2. AnonymousJanuary 23, 2010 at 9:32 AM

    AB=BCtan40
    BD=ABtan20
    EB=BCtan10
    tan(EDB)=tan10/(tan20.tan40)=tan30
    x=70-30=40
    .-.

    ReplyDelete
  3. AnonymousJanuary 26, 2010 at 5:39 AM

    angle B + angle A + angle C =180
    90 + 50 + angle ECA + 10 = 180
    hence, angle ECA = 30
    now,
    angle A + angle ECA + angle AEC = 180
    50 + 30 + angle AEC = 180
    angle AEC = 100
    now,
    angle EAD + angle AED + angle ADE = 180
    20+100+angle ADE(or x)= 180
    hence,
    x = 60

    ReplyDelete
  4. Antonio GutierrezJanuary 26, 2010 at 6:02 AM

    To Anonymous:
    The answer is not correct.

    ReplyDelete
  5. AnonymousJanuary 26, 2010 at 1:58 PM

    Why AB=BCtan40. Anybody can help.

    ReplyDelete
  6. AnonymousMarch 19, 2010 at 10:55 PM

    Set A be Centre,
    CED=30/2=15
    CED+x=60
    x=45

    ReplyDelete
  7. Antonio GutierrezMarch 20, 2010 at 8:51 AM

    To Anonymous (March 19, 2010 10:55 PM):
    The answer is not correct.

    ReplyDelete
  8. AnonymousAugust 18, 2014 at 9:57 AM

    Let circumcircle of EDC meet DA at F.
    Triangle FEC forms a trigonometric cevian model.
    From angle C in clockwise, we get 60-x,10,10,100,30,x-30.
    This is the classical 10-10-10-20-100-30 problem which makes x-30=10 => x=40.

    ReplyDelete
  9. Sumith PeirisDecember 29, 2015 at 4:36 AM

    Pure Geometry proof

    AD, CE meet at F. Since AF = CF let MF the perpendicular bisector of AC meet BC at G and AB extended at H

    Now BC is the perpendicular bisector of EH and since it therefore bisects < ECH and DF bisects < CFH (= 120) D is the incentre of Tr. FHC

    Hence <DHG = 20, < BDH = BDE = 30 and so x must be < 40 since < ADB = 70

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  10. APOSTOLIS MANOLOUDISAugust 16, 2016 at 1:01 AM

    Problem 421
    The circle (D, DC) intersects the AC, CE in K, L respectively then <LDK=2<LCK=2.30=60.So
    Triangle KDL is equilateral .Is <DLC=<DCL then <ELK=<EAK=50 so AKEL is cyclic. But <KLD=60=
    2.30=2<KAD therefore the K is circumcenter the triangle AKD so LA=LK=LD=KD=DC and
    <LAK=<LKA=50+30=80 or <LKE=<LAE=80-50=30.Then triangle LKE=triangle DKE so LE=ED and
    <EDL=10,so <EDA=<BDA-<BDE=70-(20+10)=40.
    Αnother solution the circle (E,EA) similar.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  11. AnonymousMay 2, 2020 at 8:12 PM

    It is special case of more general problem: (30-2t, 120-3t, 60-2t, 30) where t = 10.

    The problem is related with following cevian problem.
    It can be generated by taking P on EC where AEDP concyclic.
    https://output.jsbin.com/fofecum#100,10,10,30

    ReplyDelete
  12. Mohamed GAROUAOUISeptember 12, 2024 at 6:37 AM

    Hi,this is my approach:
    extend from C line FC such that CF=FA
    for notation , let CF=a
    since AFC is isocelles, therfore <FCB=10°
    we conclude that CEF is isocelles (since CB is bisector and perpondicular to EF)
    so : CE=CF=AF=a
    we have also ED=DF (since D is on BC) (1)
    now draw equilateral triangle CFH so we have :
    FH=HC=FC=a
    note that A,G and H are colinear ,because <EHA=180°-20°-60°-80°=10°
    and since FH=FA therfore FAH is isocelles
    note also that <HDC=180°-70°-40°=70°
    thus,HD=HC=FH=a
    since <HFC=60°
    therfore: <DFC=20°
    and then we have:<EFD=60°
    by the result (1) we have so far triangle EFD is equilateral
    thus we have finaly:
    x+60°+80°=180°
    therfore:
    x=40°

    ReplyDelete

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