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.
This entry contributed by Erina-NJ.
Details: Click on the figure below.

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

1 comment:

  1. 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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