Center of the given circle is O.

At Q,the tangent at P meets.

Then join BP.

Required to prove: $BQ=QC$

Proof :

$\angle ABC={{90}^{\circ }}$
In $\vartriangle ABC,\angle 1+\angle 5={{90}^{\circ }}$
And, $\angle 3=\angle 1$

[angle between chord and tangent the chord equals angle made by the chord]
So,

$\angle 3+\angle 5={{90}^{\circ }}$  ……..(i)

Also, $\angle APB={{90}^{\circ }}$  [angle in semi-circle]
$\angle 3+\angle ={{90}^{\circ }}$  …….(ii) [∠APB + ∠BPC = 180°]
From (i) and (ii), we get

$\angle 3+\angle 5=\angle 3+\angle 4$

$\angle 5=\angle 4$

$\Rightarrow PQ=QC$
Also, $QP=QB$

[tangents drawn from an inner point to a circle are equal]
$\Rightarrow QB=QC$

– Hence proved.