Question

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

• A. True
• B. False

Answer 1

The correct option is A True

$△ABC$ is right angled triangle.

$\angle ABC={90}^o$

A circle is drawn on side $AB$ as diameter intersecting $AC$ in $P,$ $PQ$ is the tangent to the circle when intersects $BC$ in $Q$

Now, join $BP$

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

Length of tangents drawn from an external point to the circle are equal

$\therefore$ $PQ=BQ$ ----- ( 1 )

$\Rightarrow$ $\angle PBQ=\angle BPQ$ [ Equal sides have equal angles opposite to them

$AB$ is a diameter.

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

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

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

$\therefore$ $\angle BPC={90}^o$

In $△BPC,$

$\Rightarrow$ $\angle BPC+\angle PBC+\angle PCB={180}^o$

$\Rightarrow$ $\angle PBC+\angle PCB={180}^o-\angle BPC={180}^o-{90}^o$

$\therefore$ $\angle PBC+\angle PCB={90}^o$ ------ ( 2 )

Now, $\angle BPC={90}^o$ ----- ( 3 )

From ( 2 ) and ( 3 ), we have

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

$\Rightarrow$ $\angle PCQ=\angle CPQ$ [ Since, $\angle BPQ=\angle PBQ$ ]

In $△PQC,$

$\Rightarrow$ $\angle PCQ=\angle CPQ$

$\therefore$ $PQ=QC$ ----- ( 4 )

From ( 1 ) and ( 4 ), we have

$\Rightarrow$ $BQ=QC$

$\therefore$ Tangent at $P$ bisects the side $BC$