Let and .

If is cyclic, then simple angle chasing gives and . Also, and , so
and
so as well. Thus and .

Now we assume . We can use the Law of Sines on , , , to obtain

\[\begin{align}
\frac{AP}{\sin\angle{DBC}}=\frac{AP}{\sin\angle{ABP}}=\frac{AB}{\sin\angle{APB}}=\frac{AB}{\sin\angle{APY}}\\
\frac{AP}{\sin\angle{BDC}}=\frac{AP}{\sin\angle{ADP}}=\frac{AD}{\sin\angle{APD}}=\frac{AD}{\sin\angle{APX}}\\
\frac{CP}{\sin\angle{DBA}}=\frac{CP}{\sin\angle{CBP}}=\frac{CB}{\sin\angle{CPB}}=\frac{CB}{\sin\angle{CPY}}\\
\frac{CP}{\sin\angle{BDA}}=\frac{CP}{\sin\angle{CDP}}=\frac{CD}{\sin\angle{CPD}}=\frac{CD}{\sin\angle{CPX}}.
\end{align}\]

From (1)(3)/(2)(4), we have

\[\begin{align*}
\frac{\sin\angle{BDC}}{\sin\angle{DBC}}\cdot\frac{\sin\angle{BDA}}{\sin\angle{DBA}}
&=\frac{AB}{AD}\cdot\frac{CB}{CD}\cdot\frac{\sin\angle{APX}}{\sin\angle{APY}}\cdot\frac{\sin\angle{CPX}}{\sin\angle{CPY}}\\
&=\frac{AB}{AD}\cdot\frac{CB}{CD}\cdot\frac{\cos(\angle{APX}+\angle{CPX})-\cos(\angle{APX}-\angle{CPX})}{\cos(\angle{APY}+\angle{CPY})-\cos(\angle{APY}-\angle{CPY})}.
\end{align*}\]

But by the Law of Sines on and , we know
Multiplying this with the previous equation we now have
Since (i.e. does not lie on or else would lie on and bisect ), this means that . Using , (2)/(3) now gives us
Thus the circumradii of and are equal, and we're done since is convex.*

I feel so crappy for not realizing that were isogonal conjugates with respect to .

Edit: Alex pointed out that this is obviously incorrect. Easy angle chasing eventually kills this though.