Let $ A$ and $ B$ be the intersections of two circumferences $\Gamma_1$, and $\Gamma_2$. Let $C$ and $D$ points in $\Gamma_1$ and $\Gamma_2$ respectively such that $AC = AD$. Let $E$ and $F$ be points in $\Gamma_1$ and $\Gamma_2$, such that $\angle ABE = \angle ABF = 90^o$. Let $K_1$ and $K_2$ be circumferences with centers $E$ and $F$ and radii $EC$ and $FD$ respectively. Let $T$ be a point in the line $AB$, but outside the segment, with $T\ne A$ and $T \ne A'$, where $A'$ is the point symmetric to $A$ with respect to $ B$. Let $X$ be the point of tangency of a tangent to $K_1$ passing through $T$, such that there arc two points of intersection of the line $TX$ to $K_2$. Let $Y$ and $Z$ be such points. Prove that $$\frac{1}{XT}=\frac{1}{XY} + \frac{1}{XZ}.$$