In a right triangle ABC in which B = , 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.
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
[ Sum of interior angle of a triangle is 180°]
Also [ tangent and chord made equal angles in alternate segment]
[ angle in semi circle formed is 90°]
Equal equation (i) and (ii) we get
[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