Question

In a right triangle $ABC$ in which $\angle B=90°$, 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

Step 1: Drawing the diagram

Given: $AB$ is the diameter of the circle.

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

Let's draw the figure.

From the figure,

$PQ=BQ$ …………$\left(1\right)$ [ Length of tangents drawn from an external point to the circle are equal ]

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

$\angle APB=90°$ [ Angle in a semi-circle is right angle ]

$\angle APB+\angle BPC=180°$ [ Linear pair ]

$$\begin{gathered} \angle BPC=180°-90° \\ \angle BPC=90° \end{gathered}$$

Step 2: Prove the statement

In$∆BPC,$

$\angle BPC+\angle PBC+\angle PCB=180°$ [ Angle sum property ]

$\angle PBC+\angle PCB=180°-90°$

$\angle PBC+\angle PCB=90°$……………$\left(2\right)$

Now,

$\angle BPC=90°$

$\angle BPQ+\angle CPQ=90°$……………$\left(3\right)$

From, equation $\left(2\right)$ and equation $\left(3\right)$, we get

$$\begin{gathered} \angle PBC+\angle PCB=\angle BPQ+\angle CPQ \\ \Rightarrow \angle PCQ=\angle CPQ \end{gathered}$$ $\left[\because \angle BPQ=\angle PBQ,\angle PCB=\angle PCQ,\angle PBQ=\angle PBC\right]$

In $∆PQC,$

$\angle PCQ=\angle CPQ$

$\Rightarrow PQ=QC$……………$\left(4\right)$

From, equation $\left(1\right)$ and equation $\left(4\right)$, we get

$BQ=QC$

Thus, the tangent at $P$ bisects the side $BC.$