How to show it is a rhombus I am trying to solve question 2 (figure 2). I have shown that the diagonals are interesting each other in right angle but I cannot show that AB||GH. Please help.

Using Alternate Interior Angles
$$\angle GBH=\angle AHB$$ and  $$\angle BHG=\angle ABH$$
Again, $$\angle ABH=\angle GBH$$
$$\implies\angle ABH=\angle AHB\implies AB=AH\ \  \  \ (1)$$
Similarly $BG=GH$
In $\triangle ABH, \triangle GBH$  $$\angle ABH=\angle GBH,\angle AHB=\angle GHB$$
and $BH$ being the common side
Using SAA Congruence
$$\triangle ABH\cong\triangle GBH$$
$$\implies AB=BG,AH=GH$$
Using $(1),$  $$BG=AB=AH=GH$$
