Turkey 2000-2

[Masum]

Let $O$ be the centre of a circle having two tangents $SP,SQ$ from a point $S$ outside of the circle and join $S,O$ such that it intersects the circle at $A,B$ with $B$ nearer to $S$.$X$ is a point in the minor arc $PB$.$PX$ and $QX$ intersect $SB$ at $D,C$ respectively.
Prove that $\frac 1 {AC}+\frac 1 {AD}=\frac 2 {AB}$

[Zzzz]

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Why have you posted this here ?

[Moon]

The user has been warned. Failing to follow the rule further will result in something that I don't want to do.

[Navedimtiaz]

As
$SP=SQ$ ,
$BA$ is the diameter,
so by symmetry we get arc $PB=$ arc $BQ$
so, $\angle BQP$ $=$ $\angle BXQ$
But,
$\angle DXB$ $=$180-$\angle PXB$ = $\angle BQP$
So.
$BX$ is the internal angle bisector of $\angle DXC$
hence,$\frac{CB}{BD} $ = $\frac{CX}{XD}$
again.arc$PA$=arc$QA$ .so, $\angle PXA$ $=$ $\angle QXA $
So. $XA$ is the external angle bisector of $\angle DXC$
hence, $\frac{AC}{AD}=\frac{CX}{XD}$
That means,$\frac{CB}{BD}=\frac{AC}{AD}$
$\Rightarrow \frac{CB}{AC} =\frac{DB}{AD}$
$\Rightarrow $ $1$ $+$ $\frac{CB}{AC}$ $+$ $1$ $-$ $\frac{DB}{AD}= 2$
$\Rightarrow $ $\frac{AB}{AC}+\frac{AB}{AD}= 2$
So,the statement is proved........

[Masum]

Let $PC$ intersect arc $QB$ at $Y$, arc $BX=BY=>$ $\angle CPB=\angle YPB=\angle BPX=\angle BPD=>$ $BP$ bisects angle $CPD$ internally.
Note that $\angle APB=90=>PA$ bisects $\angle CPD$
So $\frac {AC} {AD}=\frac {BC} {BD}=>\frac 1 {AC}-\frac 1 {AB}=\frac 1 {AB} -\frac 1 {AD}$.
Q.e.d.