Is it a Trapezoid? I have a convex quadrilateral (ABIJ) similar to a trapezoid. 

By condition $ABCDEFGHIJ$ is a regular decagon. 
Then $AB = JI$, $\angle BAJ = \angle IJA$.
We can prove that $\angle IBA= \angle BIJ, AI = BJ$.
Is it enough to say that $ABIJ$ is a trapezoid? How to verify AJ || BI ?  
 A: Call the intersection point of $BJ$ and $AI$, $K$. We have that $$\triangle IBJ \cong \triangle BIA \hspace{1.5 cm} \text{(by SSS criterion)} $$ and so $\angle IBJ=\angle BIA = a$.
Also, similarly  $$\triangle BJA \cong \triangle IAJ$$ and so $\angle BJA =\angle IAJ =b$. 
In $\triangle BKI$, $\angle BKI =\pi-2a$ and in $\triangle AKJ$, $\angle AKJ=\pi-2b$ and by vertically opposite angles we have that $$\angle BKI=\angle AKJ \\ \implies a=b$$ which further implies by reasoning of alternate angles being equal that $$AJ \ || \ BI$$
A: 
Let $\angle IBJ=x$ & $a$ be each side of regular decagon. We have
$$\text{Using sine rule in} \ \Delta ABJ, \ \ \ \frac{BJ}{\sin 144^\circ}=\frac{a}{\sin 18^\circ}\implies BJ=\frac{a\sin36^\circ}{\sin 18^\circ}$$
$$\text{Using sine rule in} \ \Delta IBJ, \ \ \ \frac{BJ}{\sin (54^\circ-x)}=\frac{a}{\sin x}\implies BJ=\frac{a\sin(54^\circ-x)}{\sin x}$$
Equating the values of $BJ$, $$\frac{a\sin(54^\circ-x)}{\sin x}=\frac{a\sin36^\circ}{\sin 18^\circ}\iff x=18^\circ\ \ (0<x<90^\circ)$$
Since alternate angles are equal i.e. $\angle IBJ=\angle AJB=18^\circ$ hence $AJ \parallel BI$. This proves that $ABIJ$ is a trapezoid. 
