Question

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

$△ABC$ is a right angled triangle

$\angle ABC={90}^o$

A circle is drawn with diameter intersecting $AC$ in $P$, $PQ$ is the tangent to circle which intersects $BC$ at $Q$.

Join $BP$

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

$\therefore$ $PQ=BQ....\left(1\right)$ (length of tangents drawn from an external point to the circle )

$\angle PBQ=\angle BPQ$ (In a triangle, angle opposite to equal sides are given that $AB$ is the diamter

$\angle APB={90}^o$ (angle in a semicircleis a right angle)

$\angle APB+\angle BPC={180}^o$ (linear pair)

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

Consider $△BPC$

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

$\therefore$ $\angle PBC+\angle PCB={180}^o-\angle BPC={180}^o-{90}^o={90}^o....\left(2\right)$

$\angle BPC={90}^o$

$\therefore \angle BPQ+\angle CPQ={90}^o$ ...(3)

from 2 and 3

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

$\angle PCQ=\angle CPQ$ (Since $\angle BPQ=\angle PBQ)$

Consider $\angle PQC$

$\angle PCQ=\angle CPQ$

$\therefore$ $PQ=CQ...\left(4\right)$

From (1) and (4) we get

$BQ=QC$

$\therefore$ tangent at $P$ bisect the side $BC$