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Such a pentagon cannot exist.

Suppose there is a pentagon, ABCDE.

A line of symmetry must pass through a vertex and the midpoint of the opposite side - otherwise there will be more vertices on one side than on the other.

Suppose a line of symmetry passes through A.

Then AB = AE and BC = ED

and ÐB = ÐE and ÐC = ÐD.

Now the second line of symmetry can pass through an adjacent vertex (B or E) or a non-adjacent vertex (C or D). By symmetry, the choice between B and E is irrelevant so suppose the line of symmetry is though B.

Then BA = BC and AE = CD.

and ÐC = ÐA and ÐD = ÐE.

Combining this with the earlier result, all five sides are of equal measure, as are all five angles and so the pentagon is regular and has 5 lines of symmetry.

Alternatively, if the second line of symmetry passes through C, then

CB = CD and BA = EA

and and

ÐB = ÐD and ÐA = ÐE.

So, again, combining this with the first result, all five sides are of equal measure, as are all five angles and so the pentagon is regular and has 5 lines of symmetry.

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