Here $\angle$ABD and  $\angle$ABC are supplementary, since they lie on a straight line.
Hence $\angle$DBC = 180${}^{\circ }$
Given, $\angle$ABC = 90${}^{\circ }$
$\therefore \angle$ ABD = 180 - $\angle$ ABC = 180 - 90 = 90${}^{\circ }$

Let the two angles be $\angle ABC$ and $\angle PQR$.
There are two possible cases:

i)

 
Here, 
$\angle 1=\angle 2$ [Correspoding angles] and
$\angle 2=\angle 3$ [Corresponding angles]
$\therefore \angle 1=\angle 3$
$\Rightarrow \angle ABC=\angle PQR$

ii)

Here $\angle 1+\angle 2={180}^{\circ }$ [Co-interior angles] and
$\angle 2=\angle 3$ [Corresponding angles]
$\therefore \angle 1+\angle 3={180}^{\circ }$
$\Rightarrow \angle ABC+\angle PQR={180}^{\circ }$
Hence it can be seen that the angles could either be equal or supplementary.

Since AB || CD

∠APQ = ∠PQR (Alternate interior angles of transversal PQ)

⇒x = 50${}^{\circ }$

∠APR = ∠PRD (Alternate interior angles of transversal PR)

⇒y + 50${}^{\circ }$ = 127${}^{\circ }$

⇒ y = 127${}^{\circ }$ - 50${}^{\circ }$

= 77${}^{\circ }$