How to show it is a rhombus

Question

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.

Answer 1

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.

Answer 2

$\angle BAH+\angle ABG = 180^\circ$ 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.

Answer 3

$\angle BAO=\angle HAO=\angle BGO$, 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 $GH\|BA$.

Answer 4

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 $(1),$ $$BG=AB=AH=GH$$

Answer 5

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 $AB||GH$.