Minimizing Perimeter of a quadrilateral Let us say that we are given a quadrilateral where the diagonals are congruent and fixed at a certain length, and the angle between the two diagonals are fixed. How would you prove that the minimum perimeter is achieved when the quadrilateral is a rectangle? I know it when diagonals aren't fixed at a certain length, one can prove it is a square by considering consecutive sides. However, when the length isn't convenient, and on top of that FIXED, the conditions are different, and you cannot do the same argument. Any help on how to convince/prove to me that the rectangle minimizes the perimeter? 
 A: I just realized that if, in a parallelogram, the diagonals have the same length (I think this is what you mean by congruent), it is necessarily a rectangle !
Maybe, I haven't well understood your question. In this case, could you made it more explicit by going directly to the question without explaining the origin of it (which confuses the reader) ?
A: Call $a,b,c,d$ the vertices and view them as vectors in $\mathbb R^2$. Call $m,n$ the midpoints of the two diagonals $\overline{ac}$ and $\overline{bd}$. Finally, suppose that the baricenter $(a+b+c+d)/4$ is the origin. Then it is easy to check that $m=-n$ (because the baricenter coincides with the midpoint between $m$ and $n$). In this way you can write
\begin{align}
a&=m+v\\
b&=-m+w\\
c&=m-v\\
d&=-m-w
\end{align}
where $2v=a-c$ and $2w=b-d$ are the diagonals. Now
\begin{align}
&\overline{ab}=|a-b|=|2m+(v-w)|\\
&\overline{bc}=|b-c|=|2m-(v+w)|\\
&\overline{cd}=|c-d|=|2m-(v-w)|\\
&\overline{da}=|d-a|=|2m+(v+w)|
\end{align}
and summing the two opposite pairs we obtain by triangle inequality that
$$\overline{ab}+\overline{cd}=|2m+(v-w)|+|2m-(v-w)|\geq 2|v-w|$$
and
$$\overline{bc}+\overline{da}=|2m-(v+w)|+|2m+(v+w)|\geq 2|v+w|,$$
and summing
$$Per(abcd)\geq 2(|v+w|+|v-w|).$$
Since the length of the diagonals is the same and the angle between them is fixed, then $2(|v-w|+|v+w|)$ is fixed and equal to the perimeter of the rectangle with that diagonals. Moreover this is basically the only case, because the equality case holds in both inequalities if and only if $\{m,v-w,v+w\}$ are linearly dependent, and that's not the case unless $v,w$ are linearly dependent, which corresponds to the trivial case where the angle is zero and can be dealt with separately. 
