Question

In a right triangle $ABC$ in which $\angle ={90}^o$, 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

PQ and BQ are tangents drawn from an external point Q.

$PQ=BQ...................\left(i\right)$ [Length of tangents drawn from an external point to the circle are equal]

$\angle PBQ=\angle BPQ$ [In atriangle, equal sides have equal angles opposite to them]

As, it is given that

AB is the diameter of the circle

$\therefore \angle APB={90}^o$ [Angle in a semi-circle is right angle]

$\angle APB+\angle BPC={180}^o$ [Linear pair]

$\angle BPC=180-90={90}^o$

In $△BPC$

$\angle BPC+\angle PBC+\angle PCB={180}^o$ [Angle sum property]

$\angle PBC+\angle PCB=180-90={90}^o..................\left(ii\right)$

Now,

$\angle BPC={90}^o$

$\angle BPQ+\angle CPQ={90}^o..............\left(iii\right)$

From (ii) and (iii), we get

$\angle PBC+\angle PCB=\angle BPQ+\angle CPQ$

$\angle PCQ=\angle CPQ$ [$\because \angle BPQ=\angle PBQ$] $[\angle PCB=\angle PCQ,\angle PBQ=\angle PBC]$

In $△PQC$

$\angle PCQ=\angle CPQ$

$\therefore PQ=QC................\left(iv\right)$

From (i) and (iv), we get

$BQ=QC$

Thus, tangent at P bisects the side BC.