## Saturday, December 31, 2011

### Problem 710: Parallelogram, Midpoint, Diagonal, Collinear Points, Metric Relations, Triangle

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

Click the figure below to see the problem 710 details.

1. 1. AC bisect BD , CO is a median of triangle BCD
So CD will pass through center of gravity N of triangle BCD and A,O,P,N, C are collinear and EF//BD
2. Solution of problem 709 showed that AG/GE=AH/AF = 4 so GH//EF//BD
In triangle BCD, since O is midpoint of BD so P is the midpoint of MQ
In triangle BND, since O is midpoint of BD so P is the midpoint of GH
So PG=PH=HQ=GM
3. see item 2 above
4. in quadrilateral MBDR, BD//MR and BM//DR so MQDR is a parallelogram and BD=MR

2. Additional clarifications for question 2 ( MG=GP=PH=HQ=QR )
Let BD=x and BD cut AE and AF at S and T
1. In tri. ABC , S is the centroid so BS=2/3 BO=x/3
Similarly DT=x/3
2. In tri. BND , since NO is a median so P is the midpoint of GH
Similarly in tri. BCD , P is the midpoint of MQ
3. Tri BGS~Tri. FGE so BG/GF=BS/EF=2/3  BG/BF=2/5  GM=x/5
FG/GB=3/2  FG/FB=3/5 = GH/BT GH=3/5* 2x/3=2x/5  PQ=1/2GH=x/5=PH
PQ=PM=GM+1/2GP=2x/5 HQ=PQ-1/GH=x/5
4. BDRM is a parallelogram  MR=BD=x  QR=MR-MQ=x/5
So MG=GP=PH=HQ=QR=x/5