Saturday, July 12, 2014

Geometry Problem 1028: Right Triangle, Double Angle, Midpoint, Congruence

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1028.

Online Math: Geometry Problem 1028: Right Triangle, Double Angle, Midpoint, Congruence

Posted by Antonio Gutierrez at 12:47 PM
Labels: , , , , , ,

10 comments:

  1. UnknownJuly 13, 2014 at 4:09 AM

    Solution:
    1)Connect B and E Notice that Triangle ABE is isosceles.
    2)Let AE = AB = a, let DE = BE = EC = b(right triangle).
    3)AEB ~ DBE (They both isosceles and have the same angles). From here we get that AE/BE = BE/DB =
    a/b = b/DB --> DB = b^2/a.
    4)Draw EN_|_BC and EM _|_AC (notice that EN = DB/2 = b^2/2a).
    5)AEM ~ CEN (Same angles) --> AE/EM = EC/ EN = a / EM = b/(b^2/2a).
    6)From the proportion in step 5 we get that EM = b/2, in triangle ECM angle ECM = 30 = 90 - 4x.
    Solution : The measure of x is 15.

    ReplyDelete
  2. Peter TranJuly 13, 2014 at 5:51 PM

    Trigonometry solution:
    Connect BE and let F is the midpoint of BE
    BDE and BAE are isosceles and similar
    Let BC= 1we have EC=EB= 1/(2.cos(x)) and BH= 1/(4.cos(x))
    And AB= BF/sin(x)= 1/(2.sin(2x))
    AC= AB/cos(3x)=BC/sin(3x) => tan(3x)=2. sin(2x)
    This equation has 2 solutions: x= 0 and x=15

    ReplyDelete
  3. AnonymousJuly 15, 2014 at 8:40 PM

    Prob 1027
    We build AH perpendicular BC and DF perpendicular AB
    CH=HD=1/2DC=1/2ED
    DF=DH=1/2ED
    α =10
    X=90-4α
    X=50

    Prob 1028

    BE=ED=EC EM=EN EN =1/2 EC
    Triangle ECN < ECN =30 ECN = 90-4X -> X =15

    Prob 1029
    We note <BDF = 90
    BD perpendicular AC


    Erina New Jersey

    ReplyDelete
  4. ion raduJuly 16, 2014 at 3:53 AM

    construim AN_|_BE si EM _|_AC, N pe BE,M pe BC,avemAEB ~ DEB(EM=NE=EB=BE/2=CE/2=>ECA=30, 90-4X =30=> X =15

    ReplyDelete
    Replies
    1. AjitJune 23, 2015 at 1:33 AM

      Wonderful solution, Prof. Radu. I wish your solutions were in English, though, for Google translation is never adequate or accurate.

      Delete
  5. AnonymousJuly 16, 2014 at 7:22 PM

    Trigonometric Solution 999 and 1028 ( Because they have the same picture and same solution

    1. EF perpendicular AB ---> BC=2EF

    2. Triangle AEF EF= ABsin2X
    3 Triangle ABC BC= ABtg3X

    Tg3X=2sin2X From geometric conditions 0 <3X < 90 ---> 0 <X<30

    Equation has this solution X=15


    Erina New Jersey

    ReplyDelete
  6. AnonymousJuly 16, 2014 at 7:23 PM

    Hi Antonio

    We have 2 other geometric solutions for problems 1028 .
    If you want them I can send to you.

    Have a great night .

    Erina

    ReplyDelete
    Replies
    1. Antonio GutierrezJuly 16, 2014 at 8:10 PM

      Hi Erina,
      Great. Please send your geometric solutions. Thanks

      Delete
  7. Sumith PeirisNovember 12, 2015 at 9:00 AM

    Let AE = AB = q, BD = r and BE = DE = CE = p

    BE is tangential to ADE so p^2 = rq ....(1)

    From similar Tr.s EF altitude of Tr. AEF = qr/2p = p/2 from (1)

    So < ECA = 30 and hence x = 15

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  8. Sailendra TJuly 12, 2021 at 2:31 PM

    E is circumcenter of DBC => m(ABE)=90-X => ABE is isosceles triangle.
    Let AP be the altitude of ABE , EQ be perpendicular from E onto AC.
    APE and AQE are congruent
    => EQ=EP=BE/2=EC/2
    => EQC is 30-60-90 right triangle
    => m(QCE)=90-4X=30
    => x=15

    ReplyDelete

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