Answer:
From given Conditions we conclude that [tex]\overline{EG}[/tex] is parallel to [tex]\overline{BH}[/tex]
Step-by-step explanation:
Given: [tex]\overline{FA}[/tex] , [tex]\overline{EC}[/tex] and [tex]\overline{BD}[/tex] are medians of ΔABC.
[tex]\overline{AG}\:=\:\overline{GH}[/tex]
Option 1:
[tex]\overline{GF}[/tex] is not parallel to [tex]\overline{EB}[/tex] because on extending both segment they are intersecting at A but parallel lines never intersects.
Option 2:
[tex]\overline{EG}[/tex] is parallel to [tex]\overline{BH}[/tex] because E is mid point of [tex]\overline{AB}[/tex] and G is mid point of [tex]\overline{AH}[/tex] then according to Mid Point Theorem [tex]\overline{EG}[/tex] is Parallel to [tex]\overline{BH}[/tex].
Option 3:
[tex]\overline{BH}[/tex] is not congruent to [tex]\overline{HC}[/tex]. It is clear from figure.
Option 4:
[tex]\overline{EG}[/tex] is not congruent to [tex]\overline{GD}[/tex]. Since from above option [tex]\overline{BH}[/tex] is not congruent to [tex]\overline{HC}[/tex] then using Mid Point theorem in ΔABH and ΔAHC.
We get,
In ΔABH [tex]\overline{EG}[/tex] is parallel to [tex]\overline{BH}[/tex] & [tex]\overline{EG}=\frac{1}{2}\times\overline{BH}[/tex] ....... Eqn (1)
Similarly,
In ΔAHC
[tex]\overline{GD}=\frac{1}{2}\times\overline{HC}[/tex] ........ Eqn (2)
So, from eqn (1) & (2)
[tex]\overline{EG}[/tex] ≠ [tex]\overline{GD}[/tex]
