Question

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

(k. 

Answer 1

Step 1: Find the relation between the angles.

As we know, the angle in a semicircle is $90°$.

Therefore, $\angle APB=90°$

With the help of the property of the linear pair we get $\angle BPC=90°$.

Therefore, $\angle 3+\angle 4=90°...\left(1\right)$.

Since, it is given that $\angle ABC=90°$.

Therefore,

$$\begin{gathered} \angle ABC+\angle 1+\angle 5=180° \\ \Rightarrow \angle 1+\angle 5=90°...\left(2\right) \end{gathered}$$ {Angle sum property of a triangle}

By alternate segment theorem we have:

$\angle 1=\angle 3$

So, the equation $\left(2\right)$ becomes:

$\angle 3+\angle 5=90°...\left(3\right)$

Step 2: Prove $PQ$ bisects $BC$.

Subtract equation $\left(1\right)$ from equation $\left(3\right)$.

$$\begin{gathered} \angle 3+\angle 5-\angle 3-\angle 4=0 \\ \Rightarrow \angle 5-\angle 4=0 \\ \Rightarrow \angle 5=\angle 4 \end{gathered}$$

Since, the sides opposite to equal angles are equal. Therefore,

$PQ=QC...\left(4\right)$

Since, the tangent drawn to a circle from an external point are equal.

Therefore, $BQ=PQ...\left(5\right)$

So, from the equation $\left(4\right)$ and equation $\left(5\right)$ we get.

$BQ=QC$

Hence it is proved that $PQ$ bisects $BC$.