So ef, fg, GH and eh are the midline of triangle ABD, triangle BDC, triangle ABC and triangle ADC respectively.
So EF is parallel to AB.
FG parallel DC
GH parallel AB
Parallel DC
So EF is parallel to HG.
FG parallel EH
So the quadrilateral EFGH is a parallelogram.
(2) When the angle BAC=90 degrees, the quadrilateral EFGH is a rectangle.
Proof: Because angle BAC+ angle ABC+ angle ACB= 180 degrees.
Angle BAC=90 degrees
So angle ABC+ angle ACB=90 degrees.
Because angle ABC= angle ABD+ angle CBD
So angle ABD+ angle CBD+ angle ACB=90 degrees.
Because EF is parallel to AB (proved)
So angle ABD= angle EFD.
Because FG and DC are parallel (authentication)
So angle ACB= angle BGF
Because angle DFG= angle CBD+ angle BGF
So EFD angle +DFG angle =90 degrees.
Because angle EFG= angle EFD+ angle DFG
So EFG angle =90 degrees.
Because the quadrilateral EFGH is a parallelogram (proved)
So the quadrilateral EFGH is a rectangle.