Euclidean geometry: Given two tangents $PM$, $PN$ to a circle with center $O$, find the value $QX\cdot RY$ Given a circle with center $O$. Tangents $PM$ and $PN$ are drawn from an external point $P$ to this circle. $Q\in PM$ and $R\in PN$ are points such that $O$ is the midpoint of $QR$. $X\in PM$ and $Y\in PN$ are points such that $XY$ is tangent to the circle. Given $QR=12$, find the value of $QX\cdot RY$ and show that it doesn't depend on $XY$. 

 A: Angles are directed modulo $\pi$ to avoid configuration issues. 

Reflect $Y$ over $OP$. Then $\triangle OYR\cong\triangle OY'Q$. 
Since $YX$ and $YR$ are both tangents erected from $Y$ we have $\measuredangle XYO=\measuredangle OYR$ and from the congruence, $\measuredangle OYR=\measuredangle QY'O=\measuredangle XY'O$. So $\measuredangle XYO=\measuredangle XY'O$ implying $OYXY'$ is cyclic. 
$YY'\parallel QR$ both being perpendicular to $OP$, and implying $\measuredangle YY'X=\measuredangle OQX$. 
Now in $\triangle QXO$ and $\triangle OXY$ we have $\measuredangle QXO=\measuredangle OXY$ since $XQ$ and $XY$ both are tangents erected from $X$, and $\measuredangle YOX=\measuredangle YY'X=\measuredangle OQX$. So $\triangle QXO\sim \triangle OXY$. Similarly $\triangle ROY\sim \triangle OXY$. 
Hence $\triangle QXO\sim\triangle ROY$, implying 
$$\dfrac{QX}{OQ}=\dfrac{RO}{YR}~\implies QX\cdot RY = OQ\cdot OR=OQ^2.~\square$$
A: 
It should be clear that (1) all same color marked angle pairs are equal; (2) $\angle PXY = 2\beta$ and $\angle PYX = 2\alpha$.
Through x, we get $\alpha + \beta = \theta$.
∴ $\triangle ORY \sim \triangle XQO$ because they both are similar to $\triangle XOY$.
∴ $\dfrac {OR}{QX} = \dfrac {RY}{OQ}$
Result follows.
