Friday, November 30, 2018

Geometry Problem 1403: Quadrilateral, 90 Degree Angle, Perpendicular, Congruence

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Geometry Problem 1403: Quadrilateral, 90 Degree Angle, Perpendicular, Congruence, Tutoring.

Posted by Antonio Gutierrez at 7:14 PM
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2 comments:

  1. Pradyumna AgasheDecember 1, 2018 at 10:07 PM

    Let AN and DM intersect at point Q, then we have AQ.AN=AD^2.
    Let perpendicular drawn from B to AM meets AM at P, then we have AP.AM=AB^2=AD^2
    We get AQ.AN=AP.AM, hence Quad.MPQN is cyclic,Angle MQN= Angle MPN=90 deg.
    Since Angle BPM = Angle MPN = 90 Deg, B,P and N are collinear points and BN is perpendicular to AM.

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  2. Sumith PeirisJanuary 18, 2019 at 8:06 AM

    Let AN, DM cut at X & BN, AM cut at Y.

    AX.AN = AD^2 = AB^2 so < BNX = < ABX = < AMX since ABMX is concyclic

    Hence MNXY is concyclic and so
    < MYN = < MXN = 90

    Sumith Peiris
    Moratuwa
    Sri Lanka

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