Geometry Problem

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

Click the figure below to see the problem 710 details.

## Saturday, December 31, 2011

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

Labels:
collinear,
congruence,
diagonal,
metric relations,
midpoint,
parallelogram,
similarity,
triangle

Subscribe to:
Post Comments (Atom)

1. AC bisect BD , CO is a median of triangle BCD

ReplyDeleteSo 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

Additional clarifications for question 2 ( MG=GP=PH=HQ=QR )

ReplyDeleteLet 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