Question

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

Answer 1

Hint: Draw the figure first then use the exterior angle property in the figure thus formed. Add all the results you get and you will find the answer.
Complete step by step answer:
Let us try to draw the figure first

Given: A circle with centre O.An Arc PQ of this circle subtends \[\angle \;POQ\] at O and \[\angle \;PAQ\] at a point A at the remaining part of the circle.
To Prove: \[\angle POQ = 2\angle \;PAQ\]
Construction: Join AO and extend it to point B.
Proof:
In \[\vartriangle APO\]
\[OP = OA\] (radius)
\[ \Rightarrow \angle OPA = \angle OAP\] (angles opposite to equal sides are equal)
Also by exterior angle property we know that Exterior angle is the sum of interior opposite angles
\[\begin{array}{l}
 \Rightarrow \angle BOP = \angle OPA + \angle OAP\\
 \Rightarrow \angle BOP = \angle OAP + \angle OAP\\
 \Rightarrow \angle BOP = 2\angle OAP.............................(i)
\end{array}\]
Also in \[\vartriangle AQO\]
\[OQ = OA\] (radius)
\[\angle OQA = \angle OAQ\] (angles opposite to equal sides are equal)
Again by exterior angle property we know that Exterior angle is the sum of interior opposite angles
\[\begin{array}{*{20}{l}}
{ \Rightarrow \angle BOQ = \angle OQA + \angle OAQ}\\
{ \Rightarrow \angle BOQ = \angle OAQ + \angle OAQ}\\
{ \Rightarrow \angle BOQ = 2\angle OAQ.............................(ii)}
\end{array}\]
Adding (i) and (ii) we get,
\[\begin{array}{l}
 \Rightarrow \angle BOP + \angle BOQ = 2\angle OAP + 2\angle OAQ\\
 \Rightarrow \angle POQ = 2(\angle OAP + \angle OAQ)\\
 \Rightarrow \angle POQ = 2\angle PAQ
\end{array}\]
Hence Proved

Note: Construction was the key step of tackling this problem and also its a theorem. Remember it for further reference we use this property of the circle for solving various types of numerical and geometrical problems.