Friday, April 29, 2011

Geometry Problem 596: Quadrilateral, Right Triangle, Isosceles, Midpoint

Geometry Problem
Click the figure below to see the complete problem 596.

Geometry Problem 596: Quadrilateral, Right Triangle, Isosceles, Midpoint.

Posted by Antonio Gutierrez at 7:08 AM
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3 comments:

  1. AnonymousMay 26, 2012 at 3:24 AM

    By applying Apollonius' theorem to triangle ACD and BCF to get two equations.
    Eliminating the length of CD, CF and BF will get 2x = 20.

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  2. Peter TranJuly 19, 2012 at 4:15 PM

    http://img849.imageshack.us/img849/1084/problem596.png

    Draw lines per attached sketch
    We have ED=EB and CB=CD => CE is perpendicular bisector of BD
    ∆ EFC ≅ to ∆ EGC …..( Case SAS)
    Since E and G are midpoints of AD and CD => EG=1/2.AC=10
    So x=EF=EG=10

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  3. Sumith PeirisJanuary 29, 2019 at 12:29 AM

    BCDE is a kite, hence CE is the perpendicular bisector of BD cutting CE at G say

    Consider Tr.s EFG and ABC

    < ABC = < EGF, AB/EG = BC/FG = 2

    Hence the 2 Tr.s are similar and x =2/2 = 10

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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