Question

Prove that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.\[\]

Answer 1

Hint: Draw a diagram of the arc PQ\[\] , angle $\angle PAR$ in the circle with centre O. You will get three possible cases where the angle $\angle PAR$ makes less than ${{180}^{\circ }}$, makes a straight angle ${{180}^{\circ }}$ or more than ${{180}^{\circ }}$. Try to use the property that on the same vertex the external angle is the sum of the opposite internal angles in each case to proceed in first and third case. Use the property that a diameter subtends a right angle at on the circle for the second case .\[\]

Complete step by step answer:
Let the arc in the circle denoted as PQ and the centre of the circle be denoted as O. Now the arc can either subtend an angle less than straight angle, a straight angle or an angle more than straight angle at the centre. The straight angle is the angle with measurement of ${{180}^{\circ }}$. \[\]
Case-1: The angle subtended by the arc is less than straight
 

We can see that PQ subtends the angle $\angle POQ$ which is less than straight angle at the centre O. Let A be any point on the remaining part of the circle where the arc makes an angle $\angle PAQ$. We draw ray from A passing through O towards R. \[\]
Now in the triangle $\Delta APO$ the radii are equal. So OP=OA. We use the property of triangles that the angles opposite equal sides are equal. So the angles opposite OP and PA must be equal. In symbols, $\angle OAP=\angle OPA$. \[\]
We know from the property of triangles that the exterior angle is the sum of interior angles. We can see in triangle $\Delta APO$ that $\angle POR$ is the exterior angle of $\angle OAP\text{ and }\angle OPA$. So we get $\angle POR=\angle OAP\text{ + }\angle OPA..(1)$. \[\]
Similarly in the triangle $\Delta APO$ the radii OQ and OA are equal, OQ=OA. So the angles opposite to them will have the relation $\angle OAQ=\angle OQA$. We see $\Delta AQO$ that $\angle POR$ is the exterior angle of $\angle OAQ\text{ and }\angle OQA$. So again using the property of exterior angle we get $\angle QOR=\angle OAQ\text{ + }\angle OQA..(2)$. \[\]
 Adding equation (1) and (2) side by side we get,
\[\begin{align}
  & \angle POR+\angle QOR=\angle OAP\text{ + }\angle OPA+\angle OAQ\text{ + }\angle OQA \\
 & \Rightarrow \angle POQ=2\left( \angle OAP+\angle OAQ \right)\left( \because \angle OAP\text{ }=\angle OPA,\angle OAQ=\angle OQA \right) \\
 & \Rightarrow \angle POQ=2\angle PAQ \\
\end{align}\]
Hence it is proved for the first case. \[\]
Case-2: The angle subtended by the arc is than straight
 

Here the arc PQ makes a straight angle, $\angle POQ={{180}^{\circ }}$ at the centre. So the points P,Q,O are collinear and PQ is diameter. We know from the property of a circle that a diameter subtended a right angle at any point of the circle. So the angle made by arc PQ at A is $\angle PAQ={{90}^{\circ }}$. Then we arrive at the proof $\angle POQ=2\angle PAQ$\[\]
Case-3: The angle subtended by the arc is less than straight
 

Here the arc PQ makes an angle more than straight angle, $\angle POQ>{{180}^{\circ }}$ . We proceed similarly as case-1.
So in $\Delta APO$ we $OP=OA\Rightarrow OAP=OPA$. The exterior angle is $\angle POR$. So $\angle POR=OAP+OPA=2OAP$ and in $\Delta AQO$ we get $OQ=OA\Rightarrow \angle OAQ=\angle OQA$. The exterior angle for $\Delta AQO$ is $\angle QOR$. So $\angle QOR=\angle OAQ+\angle OQA=\angle OAQ$.
Now,
\[\begin{align}
  & \Rightarrow \angle POR+\angle QOR=2\left( \angle OAP+\angle OAQ \right) \\
 & \Rightarrow \angle POQ=2\angle PAQ \\
\end{align}\]
Hence proved.\[\]

Note: The property that external (also called exterior or turning) is the sum of OPPOSITE internal angles not the internal angle present with it at the vertex. The sum of all internal angles of a polygon with $n$sides is \[{{180}^{\circ }}\left( n-2 \right)\]. The sum of all exterior angles is ${{360}^{\circ }}$. If all the internal angles of a polygon is less than ${{180}^{\circ }}$ then we call the polygon convex.