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Q. 64.7( 6 Votes )

Let <span lang="E

Class 8th Karnataka Board - Mathematics Part I 3. Axioms, Postulates and Theorems

Answer :

Let, ∠BOX= ∠DOX = k°

According the figure,

∠BOD + ∠AOD = 180°

According the problem,

∠XOY =90° ⇒ ∠XOD + ∠YOD = 90° ⇒ ∠YOD = 90° -k°

∴ ∠BOX + ∠AOY = 90°

∠AOY = 90°- k°

∴ ∠AOY = ∠YOD

OY bisects ∠AOD

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