In the figure,from an external point $P$,a tangent $PT$ and a line segment $PAB$ are drawn to a circle with centre $O$. $ON$ is perpendicular to the chord $AB$. Prove that:
$(i) \quad PA \cdot PB = PN^2 - AN^2$
$(ii) \quad PN^2 - AN^2 = OP^2 - OT^2$
$(iii) \quad PA \cdot PB = PT^2$

(N/A) $(i)$ We know that the perpendicular from the center of a circle to a chord bisects the chord. Since $ON \perp AB$,we have $AN = BN$.
Now,$PA \cdot PB = (PN - AN)(PN + BN)$.
Since $AN = BN$,we can write:
$PA \cdot PB = (PN - AN)(PN + AN) = PN^2 - AN^2$.
$(ii)$ In the right-angled triangle $\triangle ONP$,by the Pythagorean theorem:
$OP^2 = ON^2 + PN^2 \implies PN^2 = OP^2 - ON^2$.
Substituting this into the expression $PN^2 - AN^2$:
$PN^2 - AN^2 = (OP^2 - ON^2) - AN^2 = OP^2 - (ON^2 + AN^2)$.
In the right-angled triangle $\triangle ONA$,by the Pythagorean theorem,$ON^2 + AN^2 = OA^2$.
So,$PN^2 - AN^2 = OP^2 - OA^2$.
Since $OA$ and $OT$ are both radii of the same circle,$OA = OT$.
Therefore,$PN^2 - AN^2 = OP^2 - OT^2$.
$(iii)$ From $(i)$ and $(ii)$,we have $PA \cdot PB = OP^2 - OT^2$.
In the right-angled triangle $\triangle OTP$ (where $\angle OTP = 90^{\circ}$ because the tangent is perpendicular to the radius at the point of contact),by the Pythagorean theorem:
$OP^2 = OT^2 + PT^2 \implies OP^2 - OT^2 = PT^2$.
Thus,$PA \cdot PB = PT^2$.