Saturday, July 28, 2018

Geometry Problem 1370: Equilic Quadrilateral, 120 Degrees, Congruence, Midpoint, Diagonal, Equilateral Triangle

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 1370: Equilic Quadrilateral, 120 Degrees, Congruence, Midpoint, Diagonal, Equilateral Triangle.

Posted by Antonio Gutierrez at 10:49 AM
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3 comments:

  1. Sailendra TJuly 31, 2018 at 6:11 AM

    Extend AB & CD to meet at H, where m(H)=60=m(BJC)
    BHJC is concyclic
    let m(HBJ)=m(HCJ)=x
    => m(ABJ)=180-x=m(DCJ)
    Hence ABJ and DCJ are congruent => AJ=JD -------(1)
    Let m(BAJ)=m(CDJ)=y
    Since AJ=DJ, in the triangle AJD m(JAD)=m(JDA)
    => @-y=120-@+y (@ to be treated as Alpha)
    => @-y=60=m(JAD) --------(2)
    From (1) and (2) AJD is equilateral

    ReplyDelete
  2. Sumith PeirisJuly 31, 2018 at 8:47 AM

    Triangles ABJ & CDJ are congruent SAS and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. c.t.e.oJuly 31, 2018 at 9:50 AM

    Extend AB and DC to P. Ang PBJ = ang PCJ => tr ABJ congr to tr DCJ
    => a) AJ=DJ and ang BJA = ang DJC => b) Ang AJD = 60°
    From a) and b) AJD equilateral

    ReplyDelete

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