<s>[INST] Angle bisected in a parallelogram 
$ABCD$ is a parallelogram and $ID =JB$.
the point $P$ is the intersection of $DJ$ and $IB$.
Prove that the angle $X=Y$
$a,b,α,β$ are angles.
What I realized:
$a=b$.
and if we sum up these angles.
$$180=180-(x+α)+180-(y+β)+a $$
$$\iff 180=x+y+α+β-a \iff 360+2a=2(x+y)+2(α+β)$$
Please Help.
 [/INST] It is straightforward by sine law,
$\angle PID = 180^0 - \beta, \angle PJB = 180^0 - \alpha, \angle DPI = \angle BPJ = \gamma$
So in $\triangle PID,  \frac{\sin \beta}{PD} = \frac{\sin \gamma}{ID}$
Similarly in $\triangle PBJ,  \frac{\sin \alpha}{PB} = \frac{\sin \gamma}{BJ}$
As $BJ = ID$, $ PD \sin \alpha = PB \sin \beta$
Now in $\triangle PCD,  \frac{\sin X}{PD} = \frac{\sin \alpha}{PC}$
In $\triangle PBC,  \frac{\sin Y}{PB} = \frac{\sin \beta}{PC}$
So, $\frac{\sin X}{\sin Y} = \frac{PD \sin \alpha}{PB \sin \beta} = 1$
$\therefore \angle X = \angle Y$
</s>