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.
 A: Due parallelism of $ BG,AH $ 
$$\angle GBA+\angle BAH= 180^O$$
which are on one side of the transversal cutting  the two parallels. Dividing by 2 we get
$$\angle GBA/2+\angle BAH/2= 90^O= \angle BOG = \angle BOA,$$
as sum of two interior angles in a triangle, the diagonals cut at right angles, 
and by $AAS$, triangles $ BAO,BGO$ are congruent. 
$BA=BG,$ and $ABGH$ is a rhombus.
A: $\angle BAH+\angle ABG = 180^\circ$ since $AD||BC$.  From the angle bisections, it's easy to observe that $\angle AOB=90^\circ$: in $\triangle ABO$, $\angle ABO=\angle ABG/2$ and $\angle BAO=\angle BAH/2$, so $\angle ABO+\angle BAO=(\angle BAH+\angle ABG)/2=180^\circ/2=90^\circ$.  Moreover, $\angle ABO = \angle GBO$ because $BH$ bisects $\angle B$.  Observe that $\triangle ABO \cong \triangle GBO$ since they have a common side $OB$.  Therefore, $AO=OG$.
We finish the proof with $\triangle ABO \cong \triangle AHO$: $\angle BAO= \angle HAO$ as $AG$ bisects $\angle A$.  We've already proven that the diagonals of the quadrilateral $ABGH$ intersect each other at a right angle, giving $\angle AOB=\angle AOH$.  $OA$ is the common side, so we're done.
A: $\angle BAO=\angle HAO=\angle BGO$, and all angles at $O$ are $90^\circ$. Therefore $\triangle BAO\cong\triangle BGO$, and similarly $\triangle BAO\cong\triangle HAO$. It follows that $|GB|=|BA|=|AH|$, hence $GH\|BA$.
A: 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 $(1),$  $$BG=AB=AH=GH$$
A: Note:
$$mBAG=mGAH=mBGA \Rightarrow \Delta ABG \ \text{is isosceles},$$
$$mGBH=mBHA=mHBA \Rightarrow \Delta ABH \ \text{is isosceles}.$$
Since $BG||AH$ and $BG=AH$, then $AB=GH$ and $AB||GH$.
