Friday, March 4, 2016

Geometry Problem 1196: Parallelogram, Midpoint, Metric Relations

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

Click the figure below to view more details of problem 1196.

Math: Geometry Geometry Problem 1196: Parallelogram, Midpoint, Metric Relations

Posted by Antonio Gutierrez at 2:33 PM
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4 comments:

  1. Sumith PeirisMarch 5, 2016 at 12:08 AM

    Let EF and BD meet at S. Let GS meet BF at P. P is obviously the point of intersection of the diagonals of parallelogram BCFE.

    If CD = b, then DF = b/2 and GP = b/4 by applying mid point theorem to triangle BCF

    From similar triangles GPM and CDM, CD/GP = MD/MG = 4....(1)
    From similar triangles GPH and DFH, DF/GP = GH/HD = 2....(2)

    Since DG = 15 from (1) GM = 3 and from (2) GH = 5

    So MH = GH-GM = 5-3 = 2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. APOSTOLIS MANOLOUDISMarch 5, 2016 at 3:16 AM

      If BF intersects the EC at L and GL the AD at H, BF the AB at K and the CE ,AB at
      P.Then point H is centroid the triangle BCD therefore GH/HD=1/2 or GH=5.Is GL/LH=
      BG/HK=1/3 ,DP/PH=4/3. The triangle HGD is GM/MD.DP/PH.HL/LG=1 or GM=3.Therefore
      MH=5-3=2
      APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDOLLOS GREECE

      Delete
  2. Stan FulgerApril 3, 2016 at 11:05 PM

    Nobody used the intersection AB x DG! It's so fast!

    ReplyDelete
  3. c.t.e.o.February 1, 2021 at 12:20 PM

    https://photos.app.goo.gl/APVRCpovqWWQ6DoS6

    ReplyDelete

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