Q. 105.0( 2 Votes )
By the theorem:-
The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.
∠BOC = 2∠BAC
Adding above equations, we get,
⇒ ∠AOD+∠BOC = 2(∠BAC + ∠DCA) ----------- (1)
In ΔAPC, ∠PAC + ∠PCA = ∠BPC by exterior angles property.
⇒ ∠BAC + ∠DCA = ∠BPC
If AOD and BOC are supplementary to each other,
⇒ ∠BAC + ∠DCA = 90
And from above theorem, ∠BPC = 90°
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