Proof that angle subtended by arcs in the same segment are equal
Answers
Step-by-step explanation:
Given :
An arc PQ of a circle subtending angles POQ at the centre O and PAQ at a point A on the remaining part of the circle.
To prove : \angle POQ =2\angle PAQ∠POQ=2∠PAQ
To prove this theorem we consider the arc AB in three different situations, minor arc AB, major arc AB and semi-circle AB.
Construction :
Join the line AO extended to B.
Proof :
\angle BOQ = \angle OAQ+\angle AQO∠BOQ=∠OAQ+∠AQO .....(1)
Also, in \triangle△ OAQ,
OA=OQOA=OQ [Radii of a circle]
Therefore,
\angle OAQ = \angle OQA∠OAQ=∠OQA [Angles opposite to equal sides are equal]
\angle BOQ = 2\angle OAQ∠BOQ=2∠OAQ .......(2)
Similarly, BOP=2\angle OAPBOP=2∠OAP ........(3)
Adding 2 & 3, we get,
\angle BOP +\angle BOQ = 2(\angle OAP+\angle OAQ)∠BOP+∠BOQ=2(∠OAP+∠OAQ)
\angle POQ = 2\angle PAQ∠POQ=2∠PAQ .......(4)
For the case 3, where PQ is the major arc, equation 4 is replaced by
Reflex angle, \angle POQ=2\angle PAQ∠POQ=2∠PAQ
solution