Question

to verify that the angle subtended by an arc at the center of circle is double the angle subtended at any point on remaining arc
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Answer 1

Step-by-step explanation:

Proof:To verify that the angle subtended by an arc at the center of circle is double the angle subtended at any point on remaining arc

Construction:Draw a circle, mark two points P and Q ,join those to center as shown in the figure attached

To prove:

$\angle \: POQ = 2 \angle \: PAQ$

In ∆ AOP; from external angle Theorem

$\angle PAO+ \angle OPA=\angle POB... eq1$

In ∆ QAO; from external angle Theorem

$\angle QAO+ \angle AQO=\angle QOB...eq2$

Add both eqs1 and 2

$\angle PAO+ \angle OPA+\angle QAO+ \angle AQO=\angle POB+\angle QOB$

Since OA and OP are radius,so ∆ AOP is isosceles,hence

$\angle PAO =\angle OPA$

by the same way,in ∆ QAO

$\angle QAO =\angle AQO$

So,

$\angle PAO+ \angle PAO+\angle QAO+ \angle QAO=\angle POQ$

$2\angle PAO +2\angle QAO=\angle POQ$

$2(\angle PAO +\angle QAO)=\angle POQ$

$2\angle PAQ =\angle POQ$

Hope it helps you.

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