Here, EFGH is rhombus and EG is diagonal.
In
△EFG,⇒ EF=FG [ Sides of rhombus are equal ]
⇒ ∠2=∠4 [ Angles opposite to equal sides are equal ] ----- ( 1 )
Now, EH∥FG and EG is transversal. [ Opposite sides of rhombus are parallel ]
⇒ ∠1=∠4 [ Alternate angles ] ----- ( 2 )
From ( 1 ) and ( 2 ),
⇒ ∠1=∠2
⇒ EG bisects ∠E
Now, EF∥GH and EG is transversal.
⇒ ∠2=∠3 [ Alternate angles ] ----- ( 3 )
From ( 1 ) and ( 3 )
⇒ ∠4=∠3
⇒ EG bisects ∠G
Hence, EG bisects ∠E and ∠G
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