Ιn a circle the side is bisected by the tangent Ιn right triangle $ABC$ a circle with side $AB$ as diameter is drawn to intersect the hypotenuse ac at $P$. prove that tangent to the circle at $P$ bisects the side $BC$.  
 A: We need to draw a labelled picture. Let $O$ be the centre of our circle, and let $\ell$ be the tangent line. Draw the line segment $OP$. Suppose that $\ell$ meets $BC$ at $X$. We want to show that $XB=XC$.  
We use an angle-chasing argument. Let $\alpha=\angle A$ and $\gamma=\angle C$. Of course $\alpha+\gamma=90^\circ$.
Since $OA=OP$ (they are both radii), we have $\angle OPA=\alpha$. 
Note that $OP$ is perpendicular to $\ell$. 
By subtraction if follows that $\angle CPX=180^\circ -\alpha-90^\circ=\gamma$.
Thus $\triangle CPX$ is isosceles, with $XP=XC$.
But $XB=XP$ (tangents from an external point $X$).
Thus $XB=XC$. 
A: 
Let $Q$ be the point where the tangent line at $P$ meets $\overline{BC}$.


*

*$\overline{AP}\perp\overline{PB}$ by Thales' Theorem. So, $\angle BPQ$ and $\angle CPQ$ are complementary (they make a right angle), as are $\angle PBQ$ and $\angle C$ (they're the acute angles of a right triangle).

*As tangent segments from a common point, $\overline{QB} \cong \overline{QP}$. Thus, in $\triangle BQP$, we have $\angle BPQ \cong \angle PBQ$.

*Complements of congruent angles are congruent, so $\angle CPQ \cong \angle C$. Therefore, $\overline{QC}\cong\overline{QP}\cong\overline{QB}$, and $Q$ is the midpoint of $\overline{BC}$.
