Why the angle $\alpha$ in both triangles must be the same? I've managed to understand the proof of the formula for the sum of cosines, but there is one detail which I couldn't uncover: In the following picture, Why the angle $\alpha$ in both triangles must be the same? I tried to think about corresponding angles applied to several parts of this picture but got no success.

 A: O R P N are in a circle, so angle RON and angle RPN are equal.
A: Note that $QN || OM$. Hence, you have that alternate angles $QNO$ and $NOM$ are equal and hence $QNO$ is equal to $\alpha$. Now again we have that $PN \perp ON$. So we have that angle $PNQ = 90^\circ-\alpha$. In the right angled triangle $PQN$, angle $PQN=90^\circ$ and angle $PNQ = 90^\circ-\alpha$. Since sum of all the angles of a triangle is $180^\circ$, angle $NPR=180^\circ-(90^\circ+90^\circ-\alpha)=\alpha$.
Hope this helps.
A: Let angle OPN be $x+\alpha$. It is clear that ONP is a right angle.
Then-
\begin{equation}
\beta=180-90-x-\alpha=90-x-\alpha
\end{equation}
\begin{equation}
x=90-\beta-\alpha
\end{equation}
Now take triangle OPR and let angle MON be $y$.
Then-
\begin{equation}
\beta+y=180-90-x=90-x
\end{equation}
\begin{equation}
\beta+y=90-(90-\beta-\alpha)
\end{equation}
\begin{equation}
\beta+y=\beta+\alpha
\end{equation}
\begin{equation}
y=\alpha
\end{equation}
A: The legs of $RPN$ and $NOM$ are pairewise perpendicular.  Hence the corresponding angles are equal.
A: Note:The intersection of $PR$ and $ON$ will be refered to as $A$. 
$OM \bot PR$ and $ON \bot PN$, and $<OAR \cong <QAN$, so $\Delta OAR \sim \Delta PAN$ through Angle Angle Similarity. Cooresponding angles of similar triangles are congruent, so $<ROA \cong <NPA$.
