10 points! i need filling this in for school asap!

Given : EFGH is a parallelogram.

Prove: EG bisects HF and HF bisects EG.

Since, a parallelogram has two pairs of congruent and parallel sides.

Therefore, In parallelogram EFGH,

EF≅GH and EH≅FG

Also, EF║GH and EH║FG

Since,In triangles EKF and GKH,

EF≅HG

Since,  EF║HG

⇒∠FEK ≅ ∠HGK and ∠EFK ≅ ∠ GHK ( When two parallel lines are cut by a transversal alternative interior angles are congruent )

⇒ Δ EKF ≅ Δ GKH  ( ASA congruence postulate )

⇒ EK ≅ GK  and FK ≅ HK ( CPCTC)

⇒ EG bisects HF and HF bisects EG ( Definition of bisector)

Hence proved.

if you explain the the question more correctly i might be able to answer this better

step-by-step explanation: