Question

In a right triangle ABC in which B = 90, a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Then the tangent to the circle at P bisects BC.

• A. True
• B. False

Answer 1

The correct option is A True

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

$PQ=BQ$ ------(1) [ length of tangents drawn from an eternal point to the circle are equal]

$\angle PBQ=\angle BPQ$ -------(2) [In triangle, equal sides have equal angles opposite to them]

AB is the diameter of the circle.

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

$\angle APB+\angle BPC={180}^O$ (liner pair)

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

$={180}^o-{90}^o$

$\angle BPC={90}^o$

In $△BPC$,

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

$\angle PBC+\angle PCB={90}^o$ --------(3)

Now,

$\angle BPC={90}^o$

$\angle BPQ+\angle CPQ={90}^o$ -------------(4)

from $\left(1\right)$ and $\left(4\right)$ we get

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

$\angle PCQ=\angle CPQ$ $[\angle BPQ=\angle PBQ]$

In

$△PQC$

$\therefore PQ=QC$ ----------(5)

from $\left(1\right)$ and $\left(4\right)$ we get

$BQ=PQ=QC$

$BQ=QC$

Hence, tangent at $P$ bisects the sides $BC$.