Let's look at APQCNM when distance from B to PQ is h1, distance from D to MN is h2, AM = NC = a, AP = QC = b. AP = QC and AM = CN because ABCD is a rombus and PQ and MN are perpendicular to BD. When PQ is moved up by k units, MN is moved up by k units. As we move up PQ by t units, QC and AP increase by a rt, where r is constant. This ratio r is the sane for MN (except it's negative) because we have a rombus. Hence after the shift of PQ and MN, AP and QC increase by tr units, AM and CN decrease by rt units. So, AP + QC + AM + CN is const (*) Let angle DBC be w. Then angle BDC is w. Then PQ + MN = tgw * 2 * h1 + tgw * 2 * h2 = 2 * tgw * (BD - distance between PQ and MN) = const. Hence MN + PQ is const. Adding this to (*) gets the desired result.

We're looking at half of the rhombus: $ A\to2,B\to1,D\to0$. The triangle is clearly isosceles, and the problem reduces to showing that the perimeter of the inner pentagon depends only on its width. First, it is clear that the distance $ \overline{23}+\overline{24}=\overline{23}+\overline{25}=\overline{35}=d/\sin\alpha$ is invariant. Obviously, the side of length $ d$ is invariant. Call the remaining two sides $ a,b$ ($ a$ touches point $ 3$, $ b$ touches point $ 4$). Notice the condition $ d>BD/2$ forces the particular diagram configuration of $ a,b$ on opposite halves of the triangle. Then $ a\cot\alpha + d + b\cot\alpha=\overline{01}$, so $ a\cot\alpha+b\cot\alpha = (a+b)\cot\alpha$ is invariant, so $ a+b$ also is. ... done