Sunday, January 12, 2020

Geometry Problem 1453: Two Semicircles, Cyclic Quadrilateral, Concyclic Points

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

Details: Click on the figure below.

Geometry Problem 1453: Two Semicircles, Cyclic Quadrilateral, Concyclic Points, iPad apps, Tutoring.

4 comments:

  1. Let:
    1.)
    ∠EOB = α
    ∠EQO = β
    => ∠OEQ = 180-α-β

    ____________

    2.)
    ∠OBE = ∠OEB = 90-α/2
    ∠QCE = ∠QEC = 90-β/2
    => ∠BEC = α/2 + β/2
    ____________

    3.)
    ∠EOF = γ (given)
    => ∠EQG = 360-2α-2β-γ
    ____________

    4.)
    We know:
    ∠EQC + ∠EQG = β+360-2α-2β-γ = 360-2α-β-γ = ∠GQC
    => ∠QCG = 90-∠GQC/2 = α+β/2+γ/2-90 = ∠HCB
    ____________

    5.)
    We know:
    ∠EOQ + ∠EOF = α+γ = ∠FOB
    => ∠O(C)BF = 90-∠FOB/2 = 90-α/2-γ/2 = ∠HBC
    ____________

    6.)

    From (4) and (5):
    ∠CHB = ∠BHC = ∠FHG = ∠GHF = ∠MHN = 180-α/2-β/2
    ____________

    7.)
    ∠ECB-∠HCB = ∠MCH = α/2 + β/2 + γ/2
    => 180 - ∠CMH = ∠EMH = 180 - α/2 - β/2 - γ/2
    ____________

    8.)
    ∠EBC-∠HBC = ∠NBH = γ/2
    => ∠MHC = ∠NHB = 180 - ∠CHB = α/2 + β/2
    => 180 - ∠BNH = ∠ENH = α/2 + β/2 + γ/2
    ____________

    For a Cyclic Quadrilateral is:

    (∠MEN = ∠BEC) + ∠MHN = ∠EMH + ∠ENH = 180

    ReplyDelete
  2. https://photos.app.goo.gl/eupasLBXW1f4WeJB9

    Draw altitude EP
    Let ∠ (BEP)= u and ∠ (PEC)=v
    Note that AEB and CED are right triangles
    In right triangle AEB we have ∠ (PAE)= ∠ (PEB)= u
    In circle O ∠ (PAE)= ∠ (PFE)=u
    Similarly ∠ (PEC)= ∠ (CDE)= ∠ (CGE)=v
    In triangle FHG angle( u+v) supplement to angle ∠ (FHG)
    So ∠ (CEB)=u+v supplement to ∠ (FHG) => ENHM is cyclic quad.

    ReplyDelete
  3. Let
    < CEB = u
    < FEA = < FBA = v
    < GED = < GCD = w

    So < BAF = 90 - v = < BEG = 90 - u + w and so u = v + w

    This yields < CHM = < MEN and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Let ∠FAB=x, ∠GDC=y
    Hence ∠FBA=90-x and ∠GCD=90-y => ∠CHB=MHN=x+y -----(1)
    Since AFEB is concyclic, ∠FEB=180-x=>∠GEB=x
    Similarly CEGD is concyclic, ∠CEG=180-y=>∠FEC=y
    =>∠CEB=180-(x+y) which is suppliment to (1)

    ReplyDelete