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$P$ is a point in the square $ABCD$ with side length $8$.$S$ is the minimum area of the triangles $PAB$,$PBC$,$PCD$,$PDA$,$PAC$ and $PBD$.What is the maximum of $S$?

In the answer of book I got every part of the solution except the part that "If $O$ is the point of intersection of diagonals then consider $P$ is in the triangle $AOB$ the the area of $PAB$ is smaller than the other triangles that have one same segment with sides of square"

It is clear that it will be bigger than $PDC$ but how do we prove the other two triangles?

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    The distance of $P$ from the other sides is greater than its distance from $AB$. – 2017-01-22
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    @Aretino I don't understand why? – 2017-01-22

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The distance of $P$ from side $AB$ is less than its distance from the other sides: $$ PH=QM=QK

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