MHB Divide ABCDE into two parts with equal area

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To divide pentagon ABCDE into two equal-area parts with a line through point A, the midpoint of segment CD, labeled F, can be used to draw line AF, effectively splitting the area. The discussion raises questions about whether this method applies to general convex pentagons or only specific cases. It is suggested that if M is the midpoint of segment PQ, then line AM will also achieve the desired division, provided M is positioned between points C and D. However, there is uncertainty about the necessity of this condition, as M could potentially coincide with points C or D or even lie between other segments. The conversation emphasizes the need for a clear proof to validate these geometric relationships.
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ABCDE is a pentagon,please construct a line (passing

through point A),and divide ABCDE into two parts with equal

area
 
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Is this a general pentagon, or a particular pentagon?
 
For a regular pentagon, just construct the midpoint of $CD$, call if $F$, and draw the segment $AF$. This splits the pentagon in two equal pieces.
 
Prove It said:
Is this a general pentagon, or a particular pentagon?
A general pentagon(convex)
 
Albert said:
ABCDE is a pentagon,please construct a line (passing

through point A),and divide ABCDE into two parts with equal

area
http://www.mathhelpboards.com/attachments/f28/998d1373594278-change-pentagon-into-triangle-equal-area-pentagon.jpg

Referring to Albert's beautiful solution to the problem in http://www.mathhelpboards.com/f28/change-pentagon-into-triangle-equal-area-5486/, if $M$ is the midpoint of $PQ$ then the line $AM$ will do the job, provided that $M$ lies between $C$ and $D$. I imagine that this must necessarily be the case, but I don't see how to prove it.
 
In my opinion the best proof is "a proof without words"
so again I construct a diagram and let it explain the solution
View attachment 1033
and AG is what we need as written by Opalg
"if [FONT=MathJax_Math]M is the midpoint of [FONT=MathJax_Math]P[FONT=MathJax_Math]Q then the line [FONT=MathJax_Math]A[FONT=MathJax_Math]M will do the job"
 

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Opalg said:
Referring to Albert's beautiful solution to the problem in http://www.mathhelpboards.com/f28/change-pentagon-into-triangle-equal-area-5486/, if $M$ is the midpoint of $PQ$ then the line $AM$ will do the job, provided that $M$ lies between $C$ and $D$. I imagine that this must necessarily be the case, but I don't see how to prove it.
The statement M lies between C and D is not always true
in fact M and C (or M and D)may coincide
May M also lie between D and E ?(if the length of CD is very small)
M may also lie between B and C.
(we may check this using various diagram)
 
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