Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Friday, March 4, 2016

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

Labels:
metric relations,
midpoint,
parallelogram

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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.

ReplyDeleteIf 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

If BF intersects the EC at L and GL the AD at H, BF the AB at K and the CE ,AB at

DeleteP.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

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

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