When Are the Equivalence and Inequality Met?

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SUMMARY

The discussion centers on proving the inequality \(EF + FG + GH + HE \geq 2AC\) for rectangle ABCD with points E, F, G, and H located on segments AB, BC, CD, and DA, respectively. It is established that the equality holds when E, F, G, and H are the midpoints of their respective segments. The proof utilizes geometric rearrangement and properties of midpoints to demonstrate that \(2AC = EF + FG + GH + HE\) under these conditions.

PREREQUISITES
  • Understanding of basic geometric properties of rectangles
  • Familiarity with midpoint theorem in geometry
  • Knowledge of inequalities in geometric contexts
  • Ability to interpret geometric figures and rearrangements
NEXT STEPS
  • Study the midpoint theorem and its applications in geometry
  • Explore geometric inequalities and their proofs, particularly in polygons
  • Learn about geometric transformations and their effects on lengths
  • Investigate other geometric configurations that yield similar inequalities
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Mathematicians, geometry students, and educators interested in geometric inequalities and their proofs will benefit from this discussion.

Albert1
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Rectangle ABCD ,having fours points$ E,F,G,H $ located on segments AB, BC, CD,and

DA respectively , please prove :

$EF+FG+GH+HE\geq 2 AC $

and determine when the equivalence can be taken ?
 
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Re: Ef+fg+gh+he>=2ac

suppose we have another three cards equivalent to figure 1 ,and rearranging these four cards
in a position as shown in figure 2
from figure 2 we see :2AC=AP=HQ< HG+GM+MN+NQ=EF+FG+GH+HE
figure 3 shows the equivalence will be taken when E,F,G,H are midpoints of AB,BC,CD,and DA respectively
2AC=AC+BD=EF+FG+GH+HE
View attachment 1644
 

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