Wednesday, July 4, 2018

Geometry Problem 1359: Triangle, Circumcircle, Congruent Angles, Midpoints

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 1359: Triangle, Circumcircle, Congruent Angles, Midpoints.



    Connect DA, DC, DB
    Let angle(DBA)=angle(DCA)= u
    Note that triangle (DFB) similar to triangle(DEC) … ( case AA)
    So DE/DF= DC/DB and angle(EDC)=angle(FDB) , angle (EDF)= angle (CDB)
    So triangle(EDF) similar to $(CDB) …(case AA)
    And $(CDB) is the image of $(EDF) in the spiral similarity transformation with center D, angle od rotation = angle (EDC)
    And dilation factor= DC/DE
    Median DG is the image of median DH in this transformation so $(DEC) similar to $(DHG)
    So angle (DHG)= angle (DEC)

  2. Considering usual triangle notations, Join DC and DB
    Since m(AED)=m(AFD) => ADFE are concyclic
    Also m(DEF)=m(DAF)=m(DAB)=m(DCB)
    Therefore triangles DEF and DCB are similar
    and m(EFD)=m(CBD) -----------(1)

    Connect DH and DG, which form the medians of the two similar triangles DEF and DCB
    => DH/FE=DG/BC
    => DH/FH=DG/BG --------(2)
    From (2) and (1) triangles DHF and DGB are similar
    => DF/DH=DB/DG
    => DF/DB=DH/DG -----------(3)
    From (3) triangles DFB and DHG are similar

    Hence m(DHG)=m(DFB)=m(DEC)