# 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 $XY$.

• What have you done so far? Did you at least draw a picture? – Andrei Dec 27 '16 at 19:59
• Yes I have. Can't proceed from there. – user402283 Dec 27 '16 at 20:10
• I would add after @Andrei: can you show us a picture ? – Jean Marie Dec 27 '16 at 20:16
• Added to the post. – user402283 Dec 27 '16 at 20:22
• Do we agree that OP is not in general orthogonal to OQ ? (they are on your figure) – Jean Marie Dec 27 '16 at 20:33

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 $$\dfrac{QX}{OQ}=\dfrac{RO}{YR}~\implies QX\cdot RY = OQ\cdot OR=OQ^2.~\square$$

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.