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

10 comments:

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

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

    ReplyDelete
  3. 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

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  4. To Anonymous:
    The answer is not correct.

    ReplyDelete
  5. Why AB=BCtan40. Anybody can help.

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

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

    ReplyDelete
  8. 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.

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  9. 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

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  10. 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

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