Sunday, September 28, 2014

Geometry Problem 1045: Quadrilateral, Diagonal, Triple Angle, Triangle, Congruence

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

Click the figure below to view the complete problem 1045.

Online Math: Geometry Problem 1045: Quadrilateral, Diagonal, Triple Angle, Triangle, Congruence, Polya's Mind Map.


  1. Solution by Pradyumna Agashe:
    Draw perps. BE & CF to AC & BD resply. and call the intersection of the latter two as G.
    Since CB=CD, CF bisects /_BCD and consequently /_BGE=90° - α and /_GBE = α. By the alternate segment theorem, AB²=BG*BD ---(1) and ///ly. BE²=EG*EC=EG(EG+GC)=EG²+EG*GC
    Since BCFE is concyclis, EG*GC=BG*GF and thus BE²=EG²+BG*GF= BG² - BE²+EG*GF or 2BE² =BG(BG+GF)=BG*BD/2 or 2*BE²=AB²/2 by (1). In other words, in rt. triangle ABE, BE=AB/2 which makes /_BAE = x = 30°.
    My thanks to Pradyumna for providing this excellent solution.

  2. Very nice the above solution, but we can polish it little bit: keep the same notations and additionally call M the intersection of BE and CF. See that F is midpoint of BD and E - the one of BM. Since EMFG is cyclic, we have BG*BF=BE*BM ( 1 ). Also, since AB is tangent to the circle (ADG) we get AB*AB=BG*BD=2BG*BF ( 2 ); with (1) we have AB*AB=2BE*BM=4BE*BE, or AB=2BE, getting x=30.
    Best regards,
    S. Fulger

  3. I have a different approach. For easy writing let's make alpha = a and 2 alpha = u
    Triang ACD: sin[90-(a+x}]/CD = sin[90-(2a-x)]/AC which reduces to CD/AC = cos(a+x)/cos2a-x)
    Triang ABC: sin(x)/BC = sin[180-(a+x)]/AC which reduces to BC/AC = sin(x)/sin(a+x) Note that BC = CD
    Equating both expressions: sin(a+x)cos(a+x) = sin(x)cos(2a-x) or...... sin(u+2x) = 2sin(x)cos(u-x)
    after developing and simplifying we get sin(u)cos²(x)= 3sin(u)sin²(x)
    sin(u) cancels out meaning that x is independent of alpha and finally tan²(x) = 1/3 then x - 30 deg.

  4. Let the bisector of < BCD meet BD at X and the perpendicular drawn from B to AD at Y. Let BY meet AD at Z. ( Y is outside ABCD)

    CY is perpendicular to BD and so DXZY is concyclic. So < XYZ = x = < CAB

    Hence ABCY is concyclic and since AC bisects < BCY, AB = AY which in turn implies that BZ = ZY

    So XZ = DY/2 = BD/2 do DY = BD
    But BY = DY since CY is the perpendicular bisector of BD

    So BD = DY = BY, BYD is an equilateral triangle and hence x = < BDZ = 30

    Sumith Peiris
    Sri Lanka