given: the coordinates of rhombus wxyz are w(0, 4b), x(2a, 0), yo, -4b), and z(-2a, 0).
prove: the segments joining the midpoints of a rhombus form a rectangle.
as part of the proof, find the midpoint of xy

∴ MNOP is Rectangle

midpoint of XY (N) : (a , - 2b)

Step-by-step explanation:

W (0 , 4b)   X ( 2a , 0)   Y (0 , -4b)   Z (-2a , 0)

M (midpoint of WX) : ( (0 + 2a)/2 , (4b + 0)/2)  i. e. (a , 2b)

N (midpoint of XY) : ( (2a + 0)/2 , (0 - 4b)/2)  i. e. (a , - 2b)

O (midpoint of YZ) : ( (0 - 2a)/2 , (- 4b + 0)/2)  i. e. (- a , - 2b)

P (midpoint of ZW) : ( (0 - 2a)/2 , (4b + 0)/2)  i. e. (- a , 2b)

MN: length = 2b + 2b = 4b     MN segment perpendicular to x axis (slope undefined)

NO: length = a + a = 2a     NO segment parallel to x axis (slope = 0)

OP: length = 2b + 2b = 4b     OP segment perpendicular to x axis (slope undefined)

PM: length = a + a = 2a     NO segment parallel to x axis (slope = 0)

MN = OP and MN // OP and MN ⊥ PM

NO = PM and NO // PM and NO ⊥ OP

∴ MNOP is Rectangle

midpoint of XY (N) : (a , - 2b)

please draw graph to prove

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.