Question

Question 6
In a right angle $\angle$ ABC is which $\angle B={90}^{\circ }$, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC.

Answer 1

Let O be the centre of the given circle. Suppose, the tangent at P meets BC at Q. Join BP

To Prove BQ=QC [ angles in alternate segment]
Proof $\angle$ ABC = ${90}^{\circ }$
[ tangents at any point of circle is perpendicular to radius through the point of contact]
$\therefore$ In $\Delta$ ABC, $\angle 1+\angle 5={90}^{\circ }$ [angle sum property , $\angle ABC={90}^{\circ }$]
[ angle between tangent and the chord equals angle made by the chord in alternate segment]
$\therefore \angle 3+\angle 5={90}^{\circ }$ ...(i)
Also $\angle APB={90}^{\circ }$ [ angle in semi - circle]
$\Rightarrow \angle 3+\angle 4={90}^{\circ }$ ...(ii)
[ $\angle APB+\angle BPC={180}^{\circ }$ , linear pair]
From Eqs. (i) and (ii) we get
$\angle 3+\angle 5=\angle 3+\angle 4$
$\Rightarrow \angle 5=\angle 4$
$\Rightarrow$ PQ = QC [ sides opposite to equal angles are equal]
Also, QP = QB
[ tangents drawn from an internal point to a circle are equal]
$\Rightarrow$ QB = QC
Hence proved.