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In a right triangle ABC in which \angleB = 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.

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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 \bigtriangleupABC
\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

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