Question

$\Delta ABC$ is right triangle in which $\angle ABC={90}^{\circ }$. A circle is drawn on side AB as diameter intersecting AC in P. PQ is the tangent to the circle when intersects BC in Q. Prove that tangent at P bisects the side BC.

Answer 1

Given: $\Delta ABC$ is right triangle in which $\angle ABC={90}^{\circ }$
A circle is drawn on side AB as diameter intersecting AC in P.
PQ is the tangent to the circle when intersects BC in Q.
Construction: Join BP
Proof:
PQ and BQ are tangents drawn from an external point Q.
$\Rightarrow PQ=BQ\dots \dots \left(i\right)$ [Length of tangents drawn from an external point to the circle are equal]
$\Rightarrow \angle PBQ=\angle BPQ$ [In a triangle, equal sides have equal angles opposite to them]
As, it is given that,
AB is the diameter of the circle.
$\therefore \angle APB={90}^{\circ }$ [Angle in a semi - circle is ${90}^{\circ }$]
$\angle APB+\angle BPC={180}^{\circ }$ [Linear pair]
$\Rightarrow \angle BPC={180}^{\circ }-\angle APB={180}^{\circ }-{90}^{\circ }={90}^{\circ }$
In $\Delta BPC$,
$\angle BPC+\angle PBC+\angle PCB={180}^{\circ }$ [Angle sum property]
$\Rightarrow \angle PBC+\angle PCB={180}^{\circ }-\angle BPC={180}^{\circ }-{90}^{\circ }={90}^{\circ }\dots \dots \left(ii\right)$
$\angle BPC={90}^{\circ }$
$\Rightarrow \angle BPQ+\angle CPQ={90}^{\circ }\dots \dots \left(iii\right)$
From (ii) and (iii), we get,
$\Rightarrow \angle PBC+\angle PCB=\angle BPQ+\angle CPQ$
$\Rightarrow PCQ=\angle CPQ[\because \angle BPQ=\angle PBQ]$
In $\Delta PQC$,
$\angle PCQ=\angle CPQ$
$\therefore PQ=QC\dots \dots \left(iv\right)$
From (i) and (iv), we get,
BQ = QC
Thus, tangent at P bisects the side BC