Answer:
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.
