Saturday, August 11, 2018

Geometry Problem 1378: Parallelogram, Diagonal perpendicular to a side, Distances from a Point to Vertices, Measurement

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 1378: Parallelogram, Diagonal perpendicular to a side, Distances from a Point to Vertices, Measurement.

2 comments:

  1. https://photos.app.goo.gl/oB6yLqwymZStn8EP8
    let AC meet BD at M
    FM is the median of triangles AFC and BFD
    We have AF^2+FC^2= a^2+c^2= 2( MA^2+MF^2)
    And FB^2+FD^2=b^2+d^2= 2(MB^2+MF^2)
    In right triangle ABM we have MA^2-MB^2= AB^2= e^2
    So a^2+c^2= b^2+d^2+2.e^2

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  2. Apply Apollonius to Tr.s AFC & BFD remembering that BD & AC bisect each other

    Then apply Pythagoras to ABX, X being where the diagonals intersect

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