## 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.

1. 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

1. 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

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