In a right triangle ABC in which angleB = 90degree, 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.

Name: In a right triangle ABC in which angleB = 90degree, 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.
Uploaded: Sat, 2021-01-30 14:58
Description: In a right triangle ABC in which angleB = 90degree, 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.

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

Answers (1)

Let O be the centre of the given circle.
Suppose P meets BC at point R
Construction: Joinpoints 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 segments]
$\Rightarrow \angle 4+\angle 2= 90^{\circ}\cdots \left ( i \right )$
$\angle APB= 90^{\circ}$    [ 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]