If the midpoints of the sides of an isosceles trapezoid are joined in order then is the quadrilateral formed a rhombus?

Answer 1

I'm trying to solve this as well, but it seems that my may of going about it is off, or my work is screwed up and this is what I have gotten so far.

I labeled an isosceles trapezoid as ABCD on a graph where

A=(0,0) B=(b,c) C=(d,c) and D=(a,0)

so using the midpoint formula midpt=((x1-x2)/(2)),((y1-y2)/(2)) to find the midpoint coordinates I get

AB=((b)/(2),(C)/(2)) BC=((b+d)/(2),(c)) CD=((d+a)/2),(c)/(2)) and DA=((a)/(2),(0))

Then in order to prove a quadrilateral is a rhombus you can either prove all sides are congruent or prove that the diagonal's slopes are negative reciprocals and this is where my work falls apart...

I end up getting that AB-CD=((0)/(2b-2d-2a)) and that BC-DA=((c)/(2b+2d-2a))

So I'm not really sure if my work is bad or my method but I hope this can help you solve it yourself.