Geometry Problem

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

Click the figure below to see the complete problem 828.

## Thursday, November 29, 2012

### Problem 828: Quadrilateral, Midpoint, Opposite sides, Diagonals, Concurrent lines, Triangle, Bimedian

Labels:
bimedian,
concurrent,
diagonal,
midpoint,
opposite sides,
quadrilateral,
triangle

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Connect EF,FG, GH and EH

ReplyDeleteWe have FG//BD//EH and FG=1/2. BD=EH

So EFGH is a parallelogram so EG and FH intersect at midpoint of diagonal.

Connect MF,FN,NH and MH

We have FN//CD//MH and FN=1/2.CD=MH

So MFNH is a parallelogram so diagonals MN and FH intersect at midpoint of diagonal.

So EG, FH and MN are concurrent at the midpoint O

[Vector Method]

ReplyDeleteLet Vector(BA) = u ; Vector(BC) = v ; Vector (BD) = ru + sv , where r,s are scalars.

Hence

BE = (1/2)u ;

BF = (1/2)v ;

BH = (1/2)(BA + BD) = (1/2)[(r+1)u + sv]

BG = (1/2)(BC + BD) = (1/2)[ru + (s+1)v]

BM = (1/2)(u + v)

BN = (1/2)(ru + sv)

Now The mid point of MN = (1/2)*(BM + BN) = (1/2)*(1/2)*[(r+1)u + (s+1)v]

This can be rewritten into (1/2)*{(1/2)*u + (1/2)[ru + (s+1)v]} = (1/2)*(BE+BG)

i.e. the mid point of EG.

Symmetrically, it would also be the mid point of FH.

Q.E.D.

FM es base media en tr ABC así que FM // AB.

ReplyDeleteNH es base media en tr ADB así que HN // AB.

Finalmente GM // HN.

Análogamente FN // MH.

Entonces MFNH es paralelógramo y O es el pto medio de una diagonal, luego O pertenece a MN y OM=ON.

http://www.youtube.com/watch?v=_YzAMXsARSQ

ReplyDeleteLet A,B,C,D are points of mass 1.

ReplyDeleteSince E,F,G,H,M,N are four mid-points, we have

E = 1/2(A+B)

F = 1/2(B+C)

G = 1/2(C+D)

H = 1/2(D+A)

M = 1/2(A+C)

N = 1/2(B+D)

Mid-point of EG = 1/2(E+G) = 1/4(A+B+C+D)

Mid-point of FH = 1/2(F+H) = 1/4(A+B+C+D)

Mid-point of MN = 1/2(M+N) = 1/4(A+B+C+D)

Let O=1/4(A+B+C+D)

Then EG, FH, MN are concurrent at their common mid-point O.