MHB When Are the Equivalence and Inequality Met?

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The discussion centers on proving the inequality EF + FG + GH + HE ≥ 2AC for rectangle ABCD with points E, F, G, and H on its sides. It is established that the equality holds when points E, F, G, and H are the midpoints of the respective sides of the rectangle. The proof involves rearranging segments to demonstrate that the total length of EF, FG, GH, and HE is at least twice the length of the diagonal AC. Additionally, the relationship between the segments and the diagonals is illustrated through geometric figures. The equivalence is achieved specifically when the points are positioned at the midpoints of the rectangle's sides.
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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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