Hello, djo4567!
I'll do the first one.
Connect the midpoint of the sides of a rhombus.
Prove that the midpoints form a rectangle.
$\displaystyle \text{We have rhombus }ABCD,\,\text{ and midpoints }E, F, G, H.$
$\displaystyle \text{Draw diagonal }BD.$Code:E A o * * * o * * * o B * * * * * * * * * * * * H o o F * * * * * * * * * * * * D o * * * o * * * o C G
$\displaystyle \text{In }\Delta ABD,\,E\text{ and }H\text{ are midoints of }AB\text{ and }AD.}$
$\displaystyle \text{Hence: }\,EH = \tfrac{1}{2}BD\text{ and }EH \parallel BD.$
$\displaystyle \text{(Theorem: the segment joining the midpoints of two sides of a triangle}$
. . $\displaystyle \text{is parallel to and one-half the length of the third side.)}$
$\displaystyle \text{In }\Delta CBD,\,F\text{ and }G\text{ are midpoints of }CB\text{ and }CD.$
$\displaystyle \text{Hence: }\,FG = \tfrac{1}{2}BD\text{ and }FG \parallel BD.$
$\displaystyle \text{Hence, }EFGH\text{ is a parallelogram.}$
$\displaystyle \text{(Theorem: if two sides of a quadrilateral are equal and parallel,}$
. . $\displaystyle \text{the quadrilateral is a parallelogram.)}$
$\displaystyle \text{Draw diagonal }AC.$
$\displaystyle \text{Then: }\,AC \perp BD.$
$\displaystyle \text{(Diagonals of a rhombus are perpendicular..)}$
$\displaystyle \text{In }\Delta ABC,\,EF = \tfrac{1}{2} AC\text{ and }EF \parallel AC.$
$\displaystyle \text{Since }EH \parallel BD,\,EF \perp EH.$
$\displaystyle \text{Therefore, }EFGH\text{ is a rectangle.}$