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.

Posted by Antonio Gutierrez at 2:58 PM
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3 comments:

  1. Peter TranAugust 11, 2018 at 4:22 PM

    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

    ReplyDelete
  2. Sumith PeirisAugust 11, 2018 at 11:57 PM

    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

    ReplyDelete
  3. ArifAugust 19, 2018 at 9:19 AM

    Let M be the midpoint of BD.
    Join MF and apply apollonius to the triangle BFD, then we get :

    b^2+d^2=2MF^2+BD^2/2

    which we rewrite to:

    (1).BD^2=2b^2+2d^2-4MF^2

    If we apply apollonius to the triangle AFC we can get the following equation :

    (2).AC^2=2a^2+2c^2-4MF^2

    Applying pythagoras to ABD we get :

    AD^2=AB^2+BD^2

    using equations 1 and 2 we get:

    (3).AD^2=e^2+2b^2+2d^2-4MF^2

    If we apply the parallelogram law :

    AC^2+BD^2=2AB^2+2AD^2

    using equations 1,2 and 3 :

    2a^2+2c^2+2b^2+2d^2-8MF^2=4e^2+4b^2+4d^2-8MF^2
    2a^2+2c^2=4e^2++2b^2+2d^2
    a^2+c^2=2e^2+b^2+d^2

    ReplyDelete

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