Wednesday, May 11, 2016

Geometry Problem 1213: Triangle, Circumcircle, Altitudes, Cyclic Quadrilateral, Congruence

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

This entry contributed by Peter Tran.

Click the figure below to view more details of problem 1213.

Geometry Problem 1213: Triangle, Circumcircle, Altitudes, Cyclic Quadrilateral, Congruence
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2 comments:

  1. PravinMay 11, 2016 at 8:35 PM

    ACDF is a cyclic quadrilateral and DF extends to N.
    So angle AFN = C.
    ACBM is a cyclic quadrilateral and BM extends to M.
    So angle AMN = C.
    It follows AFMN is a cyclic quadrilateral.
    Next angle ANM
    = sum of angles MNF and FMA
    = sum of angles MAF and AMF
    = angle AFE = C
    (last equality true since quad BFEC is cyclic).
    So Angles ANM and ANM are equal (each being C).
    Hence AM = AN.

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  2. Sumith PeirisMay 13, 2016 at 1:41 AM

    Let H be the Orthocentre

    < BCE = C


    = < DHC (DCEH being cyclic)


    = < AFD (BFHD being cyclic)


    = < AFN (alternate angles)….(1)





    But < AFE = C (AFEC being cyclic)….(2)


    And < AMN = C (exterior angle)….(3)


    From (1) and (3), AFMN is cylic


    Hence < ANM = C (exterior angle of cyclic quad AFMN)…(4)


    From (3) and (4), AM = AN

    Sumith Peiris
    Moratuwa
    Sri Lanka

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