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

Solution

Let O be the center of the given circle.
Suppose P meet BC at point R
Construction: Join point P and B
To Prove: BR = RC
Proof:$\angle ABC= 90^{\circ}$   [Given]
In $\bigtriangleup$ABC
$\angle ABC+\angle BCA+\angle CAB= 180^{\circ}$ [  Sum of interior angle of a triangle is 180°]
$90^{\circ}+\angle 2+\angle 1= 180^{\circ}$
$\angle 1+\angle 2= 180^{\circ}-90^{\circ}$
$\angle 1+\angle 2= 90^{\circ}$
Also     $\angle 4= \angle 1$          [  tangent and chord made equal angles in alternate segment]
$\Rightarrow \angle 4+\angle 2= 90^{\circ}\cdots \left ( i \right )$
$\angle APB= 90^{\circ}$   [  angle in semi circle formed is 90°]
$\angle APB+\angle BPC= 180^{\circ}$
$\angle BPC= 180^{\circ}-90^{\circ}$
$\angle BPC= 90^{\circ}$
$\angle 4+\angle 5= 90^{\circ}\cdots \left ( ii \right )$
Equal equation (i) and (ii) we get
$\angle 4+\angle 2= \angle 4+\angle 5$
$\angle 2= \angle 5$
$PR= RC \, \cdots \left ( iii \right )$    [Side opposite to equal angles are equal]
Also,    PR = BR         …..(iv)   [  tangents drawn to a circle from external point are equal]

From equation (iii) and (iv)
BR = RC

Hence Proved