Prove: EF = GH and EH = FG

From the given markings, the diagonals $EG$ and $FH$ bisect each other at their intersection point, let's call it $M$. This means $EM = MG$ and $FM = MH$.
Consider $\triangle EMF$ and $\triangle GMH$.

• $EM = MG$ (Given, diagonals bisect each other)
• $FM = MH$ (Given, diagonals bisect each other)
• $\angle EMF = \angle GMH$ (Vertically opposite angles) Therefore, $\triangle EMF \cong \triangle GMH$ by the Side-Angle-Side (SAS) congruence criterion.

Since $\triangle EMF \cong \triangle GMH$, their corresponding sides are equal.
Thus, $EF = GH$.

Consider $\triangle EMH$ and $\triangle FMG$.

• $EM = MG$ (Given)
• $MH = FM$ (Given)
• $\angle EMH = \angle FMG$ (Vertically opposite angles) Therefore, $\triangle EMH \cong \triangle FMG$ by the Side-Angle-Side (SAS) congruence criterion.

Since $\triangle EMH \cong \triangle FMG$, their corresponding sides are equal.
Thus, $EH = FG$.
We have proven that $\boxed{EF = GH \text{ and } EH = FG}$.

From the given markings, $EF \parallel HG$ and $EH \parallel FG$. This means $EFGH$ is a parallelogram. The diagonals $EG$ and $FH$ intersect at $T$.
Consider $\triangle EFT$ and $\triangle HGT$.

• $EF = HG$ (Opposite sides of a parallelogram are equal)
• $\angle FET = \angle HGT$ (Alternate interior angles, since $EF \parallel HG$ and $EG$ is a transversal)
• $\angle EFT = \angle GHT$ (Alternate interior angles, since $EF \parallel HG$ and $FH$ is a transversal)

Therefore, $\triangle EFT \cong \triangle HGT$ by the Angle-Side-Angle (ASA) congruence criterion.

Since $\triangle EFT \cong \triangle HGT$, their corresponding parts are equal.
Thus, $ET = TG$ and $FT = HT$.
Assuming the question meant $HT = TF$ instead of $HT = TI$, we have proven that $\boxed{ET = TG \text{ and } HT = TF}$.