Divide ABCDE into two parts with equal area

[Albert1]

ABCDE is a pentagon,please construct a line (passing

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

area

 

[Prove It]

Is this a general pentagon, or a particular pentagon?

 

[Ackbach]

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.

 

[Albert1]

Is this a general pentagon, or a particular pentagon?

A general pentagon(convex)

 

[Opalg]

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.

 

[Albert1]

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"

 

[Albert1]

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)