Saturday, September 7, 2013

Geometry Problem 919: Rhombus, Triangle, Perpendicular, 90 Degrees, Congruence

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

Click the figure below to see the complete problem 919.

Online Geometry Problem 919: Rhombus, Triangle, Perpendicular, 90 Degrees, Congruence.

Posted by Antonio Gutierrez at 1:58 PM
Labels: , , , ,

4 comments:

  1. PravinSeptember 7, 2013 at 5:37 PM

    Let a denote the length of each side of the rhombus.
    BD bisects angle ABC
    Let h be the distance of D from BA or BC.
    Clearly MG = h = HP.
    (i.e.)FG - FM = EP - EH
    So FG + EH = EP + FM

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  2. nakaSeptember 7, 2013 at 7:37 PM

    Rewrite the statement to FG-FM = EP-EN
    Both sides equals to the height of the rhombus, thus proved.

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  3. AnonymousSeptember 7, 2013 at 8:26 PM

    FM + EN = FM + EN, GM = PN from being altitude of rombus
    (GM + MF) + EN = FM + (PN + NE)
    GF + EN = EP + FM


    To Antonio: I think the problem I suggested a few days ago is more interesting that these problems. However, it is your right to make the selection.
    Erina-NJ

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