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​a circle with centre 0 circumscribes a pentagon ABCDE. If AB=BC=CD and angle BCD=126 degree Find:
1) angle AEB
2) angle AED
3) angle AOC

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Please find below the solution to the asked query:


Given:In a circumscribed pentagon ABCDE, AB=BC=CD and ∠BCD=126°To find:i. ∠AEBii. ∠AEDiii. ∠AOCIn ∆BOC and ∆COD,OC=OC    Common sideOB=OD    RadiiBC=CD     GivenSo, by SSS congruence criteria,∆BOC ≅ ∆COD⇒∠OCB=∠OCD   By CPCT⇒∠OCB=∠OCD=12∠BCD=12×126°⇒∠OCB=∠OCD=63°   .....1Now, in ∆BOC,As, OB=OC     RadiiSo, ∠OCB=∠OBC    Angles opposite to equal sides are equal⇒∠OCB=∠OBC=63°          From 1       .....2Also,∠OBC+∠OCB+∠BOC=180°       Angle sum property of triangle⇒63°+63°+∠BOC=180°             From 2  ⇒∠BOC=180°-126°⇒∠BOC=54°      .....3As, AB=BC=CD        GivenSo, ∠AOB=∠BOC=∠COD=54°               From 3      .....4i. ∠AEB=12∠AOB       Angle subtended by an arc at any point on the circle is half the angle subtended by the same arc at the centre⇒∠AEB=12×54°                From 4 ⇒∠AEB=27°ii. ∠AED=12∠AOD       Angle subtended by an arc at any point on the circle is half the angle subtended by the same arc at the centre⇒∠AEB=12×54°+54°+54°                From 4 ⇒∠AEB=81°iii. ∠AOC=∠AOB+∠BOC=54°+54°⇒∠AOC=108°

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