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Mathematics solution of grade three
(1) Proof: Because E, F, G and H are the midpoint of AD, BD, BC and AC respectively.

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.