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Question
If bisectors of angles A and D of a quadrilateral ABCD meet at 0, then show that ∠B + ∠C = 2 ∠AOD
Solution
Since AO and DO are bisectors of ∠A and ∠D of quadrilateral ABCD,
∠A = 2∠OAD and ∠D = 2∠ODA ....(i)
In ΔAOD,
∠OAD + ∠ODA + ∠ACD = 180°
⇒ 2∠OAD + 2∠ODA + 2∠AOD = 360° ....[Multiplying both sides by 2]
⇒ 2∠OAD + 2∠ODA = 360° - 2∠AOD ....(ii)
In quadrlateral ABCD,
∠A + ∠B + ∠C + ∠D = 360°
⇒ 2∠OAD + ∠B + ∠C + 2∠ODA = 360° ....[From (i)]
⇒ ∠B + ∠C = 360° - 2∠OAD - 2∠ODA
⇒ ∠B + ∠C = 360° (2∠OAD + 2∠ODA)
⇒ ∠B + ∠C = 360° - (360° - 2∠AOD) ....[From (ii)]
⇒ ∠B + ∠C = 360° - 360° - 2∠AOD
⇒ ∠B + ∠C = 2∠AOD.
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