Sunday, November 11, 2018

Geometry Problem 1399: Triangle, Circumcircle, Perpendiculars, Angles, Congruence, Cyclic Quadrilateral

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 1399: Triangle, Circumcircle, Perpendiculars, Angles, Congruence, Cyclic Quadrilateral, Tutoring.
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5 comments:

  1. Sumith PeirisNovember 11, 2018 at 5:21 PM

    Since CEFG is concyclic,
    < ECF = < EGF = x say

    Now since AGEC is concyclic,
    < EGH = < EAH = < DCB

    But < EGH = alpha + x and
    < DCB = alpha’ + x and so

    alpha = alpha’

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. Sumith PeirisNovember 11, 2018 at 5:48 PM

    Solution 2

    Since AGEH & ABDC are both concyclic
    < HEG = < BDC

    So in Tr.s BDC & HEG,
    < HEG = < BDC
    < EHG = < DBC since each angle = < CAD
    Hence < EGH = < DCB

    But < EGF = < ECF

    So alpha = alpha’

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. c.t.e.oNovember 11, 2018 at 11:56 PM

    Ang BAD = HGF + FGE ( AHEG )
    Ang BAD = BCE + ECD ( BD )
    Ang FGE = FCE ( FECG )
    => HGF = ECD

    ReplyDelete
  4. AnonymousNovember 12, 2018 at 12:46 AM

    Unghiul FEG = unghiul FCG (laturi perpendiculare).
    De aici FECG circumscriptibil
    => unghiul FGE = FCE (1).
    Unghiul AHE = unghiul AGE =90
    => AHEG circumscriptibil
    Unghiul HAE = unghiul HGE . (2)
    Dar unghiul BAD = unghiul BCD (3)
    Din (2) si (3) => unghi BCD= unghi HGE.(4)
    Din (1) si (4) => α= α^,
    Florin Popa, Comanesti, Romania

    ReplyDelete
  5. APOSTOLIS MANOLOUDISNovember 12, 2018 at 3:29 AM

    Is EFGC=cyclic=>α=<DCE=<FGH=α'.

    ReplyDelete

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