Monday, May 19, 2008

Elearn Geometry Problem 42



See complete Problem 42
Angles, Triangle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

7 comments:

  1. draw bisector of ang C
    draw ang DAE = α, E on bisector of C
    join B to D, D to E, draw altitude BH, H on DE

    tr BDE is equilateral
    ang DBH = ang DAC ( perpendicular sides )
    ang KDE = ang DAC ( AD meetBE on K, AD = EC => DE//AC)
    ang FED = ang ECA (CE meet BD on F, DE//AC )

    BH bisector of DBE, DK bisector of EDB, EF of DEB
    => DBE = 6 ∙ DAC = 180º
    => DAC = 30º

    from tr ABC
    A + C + B =( α + 30 ) + ( 30 + α ) + x = 180

    x = 120° - 2α
    -----------------------------------------------

    ReplyDelete
  2. Sir, Did u receive my solution for Problem 42?

    ReplyDelete
  3. Draw a circle having its center B and radius BA,
    Draw another circle having its center C and radius CB.
    Let K be the lower intersection point of two circles.
    Connect BK and CK
    So, Tri. BKC is an equlateral.
    Obviously ang.KBC=60deg
    Since ang.BCK=60deg, ang.DCK=60-2α.
    Since, ang.DCK is angle on center and ang.DBK is angle on circumferance subtended by same segment: ang.DBK=30-α
    similarly BCD=2.BKD
    ang.BKD=α
    now connect BD
    as per the symetrical property in triangles ADB and KBD, ang.ABD also 30-α
    so, ang.ABC=X=ang.ABD+ang.KBD+ang.KBC
    therefore X=30-α+30-α+60
    ie: X=120-2α
    ............

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  4. Drop a perpendicular from B to AD say. BE and from C to BD say CF. Tr.s DFC, BFC and ABE are all congruent ASA. So DF = BF = BE, hence Tr. BDF is 30 60 90 with < BDE = 30.

    So < ABD = 30-@ and since < FBC =90-@, x = 120-2@

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Hello, I have seen a very similar problem: B and C are known angles (ie 110 and 20), the angle to find is the one in A position.
    Never could solve that problem, I think now it will possible for me if I try to follow the comments here.
    Thank you!

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  6. Define B’ symmetric of B by AD => AB=AB’
    ang BAD = a therefore ang BAB’ = 2a
    Therefore ΔBAB’ is congruent to ΔBCD, and BB’=BD, BD=DB’
    Therefore ΔBB’D is equilateral
    Draw a line passing through B and perpendicular to AC meeting in E
    AB=BC => BE is angle bisector of ang ABC
    The drawing is symmetric, therefore BE is also angle bisector of ang DBB’
    AD is angle bisector of ang BDB’ and CB’ is angle bisector of ang BB’D
    By symmetry lines AD, CB’ and EB meet in point P
    P is on angle bisector of ang DBB’, on angle bisector ofang BB’D and on angle bisector of ang BDB’ therefore P is the center of ΔBDB’
    Therefore PD=PB=PB’
    Consider now ΔAPB
    Since ΔBB’D is equilateral, ang APB = 120
    ang APB +ang PBA +ang BAP = 180
    120+x/2+a=180 =>x+2a=120 => x=120-2a

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