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: First of all draw a circle with PQ as arc and a point A at circumference and O as center. Now, we have to prove that \[\angle POQ=2\angle PAQ\]. Now join O to A and extend it to point B. Now use the exterior angle theorem and angles opposite to equal sides are equal in \[\Delta OAQ\] and \[\Delta OAP\] to prove the desired result.

Step-by-step answer:
Here, we have to 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. Let us consider an arc PQ of a circle subtending angles POQ at the center O and PAQ at point A on the remaining part of the circle. Here, we have to prove that
\[\angle POQ=2\angle PAQ\]
Now, to prove this theorem, we consider the arc PQ in three different situations, minor arc PQ, major arc PQ, and semicircle PQ which is as follows:

Now, in the above circle, we do construction of joining A to O and extending it to B as follows.

We know that in any triangle, the exterior angle is equal to the sum of its opposite interior angles. So in \[\Delta OAQ\] of the above diagrams, we get,
\[\angle BOQ=\angle OAQ+\angle AQO.....\left( i \right)\]
Also, we know that the radius of the same circle is equal. So, we get, OA = OQ.
We know that the angles opposite to equal sides are equal. So, in \[\Delta OAQ\], we know that OA = OQ. So, we get,
\[\angle OAQ=\angle AQO.....\left( ii \right)\]
By substituting \[\angle AQO=\angle OAQ\] in equation (i), we get,
\[\angle BOQ=\angle OAQ+\angle OAQ=2\angle OAQ....\left( iii \right)\]
Similarly, now if we take \[\Delta OAP\], by exterior angle theorem, we get,
\[\angle BOP=\angle OAP+\angle OPA....\left( iv \right)\]
We know that angles opposite to equal sides are equal. So, in \[\Delta OAP\], we know that OA = OP as both are the radius of the same circle. So, we get,
\[\angle OAP=\angle OPA....\left( v \right)\]
By substituting \[\angle OPA=\angle OAP\] in equation (v), we get,
\[\angle BOP=\angle OAP+\angle OAP....\left( vi \right)\]
By adding equation (iii) and (vi), we get,
\[\angle BOP+\angle BOQ=2\angle OAQ+2\angle OAP\]
Now by the diagram, we get,
\[\angle POQ=2\left( \angle OAQ+\angle OAP \right)\]
\[\Rightarrow \angle POQ=2\left( \angle PAQ \right)\]
Also in case III where PQ is a major arc, we get,
\[\angle BOQ+\angle BOP=2\left( \angle OAQ+\angle OAP \right)\]
Reflex angle, \[\angle POQ=2\left( \angle PAQ \right)\]
\[\left( {{360}^{o}}-\angle POQ \right)=2\angle PAQ\]
Hence, we have proved that 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.

Note: In the questions involving the arc of the circle, students must take all three cases that are of major arc, minor arc, and arc as a semicircle. Also, students must remember the exterior angles theorem as it is very useful in geometry. Also, for the case of taking the arc as a major arc, we must take care of the reflex angle of \[\angle POQ\] which is \[{{360}^{o}}-\angle POQ\] while the remaining proof will remain the same for all three cases. Here, assume that we are taking \[\angle POQ\] as the smaller angle among this angle and its reflex.