Monday, May 19, 2008

Elearn Geometry Problem 45



See complete Problem 45
Angles, Triangle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 9:15 PM
Labels: ,

10 comments:

  1. AnonymousJanuary 29, 2010 at 7:27 AM

    Faltan datos, la repuesta asume que AC = CE.. y no se puede llegar a esa conclusion o en todo caso no explica como llega a ese punto.
    Espero respuestas.
    Muchas Gracias.

    ReplyDelete
  2. AnonymousNovember 14, 2010 at 4:00 PM

    http://img842.imageshack.us/img842/3018/p045angletriangle.gif

    ReplyDelete
  3. Nilton LapaMarch 20, 2012 at 8:35 AM

    Lets use a and b for alpha and beta. BD meets AC on E. Ang(ADE) = 2a+3b and ang(CDE) = a+b. Using sinus rule in ABD and BCD, we get
    AB/BD = sin (2a+3b) / sin (2a)
    and BC/BD = sin (a+b) / sin (a),
    so sin (2a+3b) / sin (2a) = sin (a+b) / sin (a).
    That equality gives
    sin (2a+3b) = 2.sin (a+b).cos (a)
    sin (2a+3b) = sin (2a+b) + sin (b)
    sin (2a+3b) - sin (2a+b) = sin (b)
    2.sin (b).cos (2a+2b) = sin (b)
    and cos (2a+2b)=1/2 or 2a+2b = 60º.
    Finally, x+2a = 90º-2b, x = 90º - (2a+2b) = 30º.

    ReplyDelete
  4. PravinMarch 21, 2012 at 8:08 PM

    Typos corrected (regret inconvenience)
    Write a for alpha and b for beta.
    Draw BE the ⊥ bisector of AC, meeting AD at F.
    BE bisects ∠ABC(=4b)
    So ∠FBC = 2b,
    which implies BD bisects ∠FBC with ∠FBD = ∠DBC = b
    Next ∆FAC is isosceles.
    So ∠FCA = x, and
    ∠FCB = ∠FAB = 2a by symmetry
    which implies ∠FCD = ∠DCB = a
    Follows D is the incentre of ∆BFC, FD bisects ∠BFC.
    ∠BFD = ∠DFC = x + x = 2x
    Denote the sum a + b by y.
    From ∆BFC we have 4x + 2a + 2b = 180° and
    from ∆ABC we have 2x + 4a + 4b = 180°
    2x + y = 90° and x + 2y = 90°
    solving which we get x = y = 30°

    ReplyDelete
  5. PravinMarch 21, 2012 at 9:56 PM

    http://s1092.photobucket.com/albums/i418/vprasad_nalluri/?action=view&current=elearngeometryproblem45.jpg

    ReplyDelete
  6. PravinMarch 23, 2012 at 5:15 AM

    Figure for e-learn geometry problem 45 to follow my proof
    Link:
    http://imageshack.us/content_round.php?page=done&l=img198/5108/elearngeometryproblem45.png#

    ReplyDelete
  7. Sumith PeirisOctober 13, 2015 at 9:46 AM

    Denote alpha by @ and beta by £

    Let the bisector of < ABC meet AD at E. Now D is easily the incentre of Tr. BCE and so < BED = < CED = x.

    Since < BAC = 2@+x = < ACB, we have 2x + 4@ + 4£ = 180 .....(1)

    Since BE is perpendicular to AC,

    x + 2@ + 2£ = 90.....(2)

    From (1) and 2 we deduce that
    @ + £ = 30 and so from (1) x = 30

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. AnonymousSeptember 11, 2019 at 5:38 PM

      Hi Sumith.
      Not important, but < BED = < CED = 2x (you made a mistake)
      Equations (1) and (2) are the same.
      From (1) and (2) ... how to deduce that @ + £ = 30 ?

      Pedro

      Delete
    2. AnonymousSeptember 11, 2019 at 6:19 PM

      Hi,
      This is my solution
      https://mega.nz/#!UhwWVI4Z!yOpCiF1X1u59Wlg0A293tGYRHKzZ-l7CXOZVQzLooLE
      Regards
      Pedro Miranda

      Delete
  8. Sumith PeirisSeptember 16, 2019 at 11:00 PM

    Dear Pedro, yes u are right, there is a mistake in my proof above

    Amended proof is as follows:-

    Let the bisector of < ABC meet AD at E.

    In Isoceles Tr. ABC 2x + 4alpha + 4 beta = 180 .. (1)

    Now D is obviously the incentre of Tr. BCE
    Hence < BED = 2alpha + 2 beta = < DEC = 2x or x = alpha + beta

    Now substituting in (1)

    2x + 4x = 180 whence x = 30

    ReplyDelete

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