Question

Topic: Circles
Class: 9th

$\sf \red{Question:}$
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.
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Answer 1

$\star \; {\underline{\boxed{\red{\pmb{\frak{ \; Given \; :- }}}}}}$

• We have been given a Circle with centre O .
• Arc AB of the Circle subtends ∠AOB and ∠APB at the point Q remaining part of Circle .

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$\star \; {\underline{\boxed{\green{\pmb{\frak{ \; To \; Prove \; :- }}}}}}$

• ∠AOB = 2 ∠APB

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$\star \; {\underline{\boxed{\orange{\pmb{\frak{ \; ProoF \; :- }}}}}}$

Here :-

${\underline{\textsf{\textbf{ In \; ∆AOP \; :- }}}}$

$\begin{gathered} \dashrightarrow \; \; \sf { AO = OP \qquad \; \bigg( Radius \; of \; Same \; Circle \bigg) } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠OPA = ∠OAP \qquad \; \bigg( Angle \; Opp. \; to \; Equal\; Sides \; are \; Equal \bigg) } \\ \end{gathered}$

So :-

$\implies \; \; {\underline{\boxed{\pmb{\sf{ ∠OPA = ∠OAP }}}}} \; {\pink{\bigstar}}$

Now :-

${\underline{\textsf{\textbf{ In \; ∆BOP \; :- }}}}$

$\begin{gathered} \dashrightarrow \; \; \sf { BO = OP \qquad \; \bigg( Radius \; of \; Same \; Circle \bigg) } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠POB = ∠PBO \qquad \; \bigg( Angle \; Opp. \; to \; Equal\; Sides \; are \; Equal \bigg) } \\ \end{gathered}$

So :-

$\implies \; \; {\underline{\boxed{\pmb{\sf{ ∠POB = ∠PBO }}}}} \; {\purple{\bigstar}}$

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${\underline{\textsf{\textbf{ By \; Exterior \; Angle \; Property \; :- }}}}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠AOQ = ∠OPA + ∠OAP } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠AOQ = ∠OPA + ∠OPA } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠AOQ = 2 ∠OPA } \\ \end{gathered}$

So :-

$\implies \; \; {\underline{\boxed{\pmb{\sf{ ∠AOQ = 2 ∠OPA }}}}} \; {\red{\bigstar}}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠BOQ = ∠BOP + ∠PBO } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠BOQ = ∠BOP + ∠BOP } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠BOQ = 2 ∠BOP } \\ \end{gathered}$

So :-

$\implies \; \; {\underline{\boxed{\pmb{\sf{ ∠BOP = 2 ∠BOP }}}}} \; {\color{darkblue}{\bigstar}}$

$\\ \qquad{\rule{150pt}{1pt}}$

${\underline{\textsf{\textbf{ Measure \; of \; ∠POQ \; :- }}}}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠POQ = 2 ∠OPA + 2 ∠BOP } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠POQ = 2 \bigg( ∠OPA + ∠BOP \bigg) } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠POQ = 2 ∠APB } \\ \end{gathered}$

So :-

$\implies \; \; {\underline{\boxed{\pmb{\sf{ ∠POQ = 2 ∠APB }}}}} \; {\orange{\bigstar}}$

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$\therefore \;$ We have Proved 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 .

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