See complete Problem 146

Varignon's Theorem: Quadrilateral, Midpoints, Parallelogram, Area, Perimeter. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Sunday, July 27, 2008

### Elearn Geometry Problem 146 Varignon s theorem and more

Labels:
area,
median,
midpoint,
parallelogram,
perimeter,
quadrilateral,
Varignon theorem

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well, to prove that EFGH is a parallelogram, one would take triangle CBD, and use the the midline theorum to get that FG is parallel to BD, then do the same for triangle ADB to get that EH is parallel to BD, and FG is parallel to BD, thus, EH and FG are parallel. Then do it again with triangles BAC and DCQA to get that HG, EF, and AC are all parallel. Now that EFGH has sides EF and GH parellel, and sides FG, and EH parallel, it is mow a parallelogram.

ReplyDeleteHow would the others be proven?

2. By the midline theorem: EF = GH = AC/2 and EH = FG = BD/2.

ReplyDeleteiii

ReplyDeletearea of trangle EBF = area of trangle FAE(as they have the same vertex and equal bases), area of trangle BAF = area of trangle FAC

so EBF = 1/4 ABC Using the sam method it can be proved that GDH = 1/4 ADC so EBF + GDH = 1/4 ABCD

iv

Using the sam method it can be proved that EAH + FCG = 1/4 ABCD

so EAH + FCG + EBF + GDH = 1/2 ABCD

by subtracting the areas EFGH = 1/2 ABCD

in triangle ABC, E and F are the midpoints of AB, AC respectively. thus by midpoint theorem we have EF=(1/2)*AC. similarly applying the theorem to triangle ADC, we get GH=(1/2)*AC. also by the same theorem we get that EF, AC and GH are parallel to each other. therefore EFGH is a parallelogram because a pair of opposite sides are equal and parallel.

ReplyDeleteEF=GH=(1/2)*AC and EH=FG=(1/2)*BD

these results are due to midpoint theorem.

adding these results we get EF+FG+GH+HE=AC+BD.

join AG. in triangleAGD, GH is the median as H is the midpoint of AD. so S4=(1/2)*area of triangleAGD. similarly in triangleACD, AG is the median and area of triangleAGD=(1/2)*area of triangleACD. so from the above relations we get, S4=(1/4)*area of triangleACD.

similarly, S2=(1/4)*area of triangleABC

S3=(1/4)*area of triangleBCD

S1=(1/4)*area of triangleBAD

S1+S2+S3+S4=(1/4)*(area of triangles ACD+ABC+BCD+BAD)=(1/4)*2*area of quadrilateral ABCD=(1/2)*area of qudrilateralABCD

it is easy to see that the area of quadrilateralABCD=S1+S2+S3+S4+S5

but S1+S2+S3+S4=(1/2)*area of quadrilateralABCD

so S5=(1/2)*area of quadrilateralABCD.

Q. E. D.

(To Antonio Gutierrez)

ReplyDeletesir,I think there's a small mistake:

EF=1/2AC=GH & EH=1/2BD=FG

so,the perimeter of EFGH=2*1/2AC+2*1/2BD=AC+BD [Not 1/2(AC+BD)]

To Aditya

Deletethe figure of problem 146 has been updated. Thanks.