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.


  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

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

  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
    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

  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
    Sri Lanka

  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!