Half secant in Circle OT and OQP are tangent and secant respectively drawn from external point $O$ of a circle centered at $C$. Mid-point M of the secant is joined to center $C$,an arc is drawn with center $O$ to be tangential at $M$ cutting normal $TC$ at $X$. Show that $TX= QM.$ ![Circle Half Secant](

$$TX^2 = OX^2-OT^2 =OM^2-OT^2 = (OQ+QM)(OP-QM)-OT^2$$ but since $OP\cdot OQ = OT^2$, $$ TX^2 = QM(OP-OQ)-QM^2 = 2 QM^2-QM^2 = QM^2 $$ as wanted.