Find the measures of the angles $x$ and $y$ in the diagram above.
I've tried using angles in the triangles and quadrilateral, exterior angles, and parallel lines. Everything I come up with reduces to $x+y=48$. How can I introduce a different equation to solve for $x$ and $y$?
On
From triangles DAC and BAC we have $\angle DAC = 65^\circ - x$ and $\angle BAC = 67-y$
Let $X$ be a point on $PC$ such that $DX=XC$. Then $\angle XDB=65-x$, which means that $XD$ is tangent to the circumcircle of $ADP$ and $XP\cdot XA=XD^2=XC^2$
Analogically, let $Y$ be a point on $PC$ such that $BY=YC$ and we have $YP\cdot YA=YB^2=YC^2$
Let's prove that $X=Y$. Actually this is obvious, since if we move a point $Z$ on the line $PC$ in direction of $P$, then $ZP\cdot ZA$ gets smaller and $ZC^2$ gets bigger. So, there is only one point $Z$ such that $ZP\cdot ZA=ZC^2$, which means that $X=Y$.
Now, from the isosceles triangle $XDB$ we have $65-x=67-y$.
Define $\angle DAC$ and $\angle BAC$ as $w$ and $z$ respectively.
Now you can write down four equations in terms of $w,x,y,z$ (working in degrees):
$$x+y=180-65-67$$ $$w+z=180-50-46$$ $$x+w=180-50-65$$ $$y+z=180-46-67$$
Can you take it from here?