6. Theorems on angle at the centre and the circumference

Explanation:

 

 

The theorem states that the angle subtended by the arc $\stackrel{⌢}{\mathit{QR}}$ of the circle at the centre $O$ is twice the angle subtended by the point $P$ at any remaining part of the circle. That is, $\angle QOR = 2\angle QPR$.

 

Proof of the theorem:

 

Consider a circle with centre $O$ and place three points $P$, $Q$ and $R$ on the circumference of the circle such that $\stackrel{⌢}{\mathit{QR}}$ subtends $\angle QOR$ at the centre $O$ and $\angle QPR$ at the circumference of the circle.

 

 

Consider three cases where (i) Figure $1$ - $\stackrel{⌢}{\mathit{QR}}$ is minor (ii) Figure $2$ - $\stackrel{⌢}{\mathit{QR}}$ is a semicircle and (iii) Figure $3$ - $\stackrel{⌢}{\mathit{QR}}$ is a major arc.

 

Extend $PO$ to $S$ and join $PS$.

 

In the triangle $POQ$, $\angle QOS$ is an exterior angle.

 

By the property of the triangle, the exterior angle is equal to the sum  of the interior opposite angles.

 

That is, $\angle QOS$ $=$ $\angle OPQ$ $+$ $\angle PQO$.

 

Here, $OP$ $=$ $OQ$. Since the radii are equal.

 

This implies that the $\angle OPQ$ $=$ $\angle PQO$ as they form the base angles of the isosceles triangle $POQ$.

 

Therefore, $\angle QOS$ $=$ $\angle OPQ$ $+$ $\angle OPQ$

 

$=$ $2 \angle OPQ$                    $……$ $\text{equation }(1)$

 

 

Add equation $(1)$ and $(2)$, we get:

 

$\angle QOS$ $+$ $\angle ROS$ $=$ $2 \angle OPQ$ $+$ $2 \angle OPR$

 

$\angle QOR$ $=$ $2 (\angle OPQ + \angle OPR)$

 

$\angle QOR = 2\angle QPR$

 

The obtained result is true for the cases (i) and (ii) whereas for case (iii) we have reflex $\angle QOR = 2\angle QPR$.