In a right triangle $ABC$ in which $\angle B = 90^{\circ}$,a circle is drawn with $AB$ as diameter,intersecting the hypotenuse $AC$ at $P$. Prove that the tangent to the circle at $P$ bisects $BC$.

(N/A) Let $O$ be the centre of the given circle. Suppose the tangent at $P$ meets $BC$ at $Q$. Join $BP$.
To prove: $BQ = QC$.
Proof: $\angle ABC = 90^{\circ}$.
In $\triangle ABC$,$\angle A + \angle C = 90^{\circ}$ (where $\angle A = \angle 1$ and $\angle C = \angle 5$).
So,$\angle 1 + \angle 5 = 90^{\circ}$ ... $(i)$
By the alternate segment theorem,the angle between the tangent at $P$ and the chord $BP$ is equal to the angle in the alternate segment,so $\angle 3 = \angle 1$.
Also,$\angle APB = 90^{\circ}$ (angle in a semi-circle).
Since $AC$ is a straight line,$\angle APB + \angle BPC = 180^{\circ}$,so $\angle BPC = 90^{\circ}$.
In $\triangle BPC$,$\angle 4 + \angle 5 = 90^{\circ}$ ... $(ii)$
From $(i)$ and $(ii)$:
$\angle 1 + \angle 5 = \angle 4 + \angle 5$
$\Rightarrow \angle 1 = \angle 4$
Since $\angle 3 = \angle 1$,we have $\angle 3 = \angle 4$.
In $\triangle PQC$,since $\angle 3 = \angle 4$,the sides opposite to equal angles are equal,so $PQ = QC$.
Also,tangents drawn from an external point to a circle are equal,so $PQ = BQ$.
Therefore,$BQ = QC$.
Hence,the tangent at $P$ bisects $BC$.