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Question
P is any point inside the triangle ABC.
Prove that: ∠BPC > ∠BAC.
Solution
Let ∠PBC = x and ∠PCB = y
then,
∠BPC = 180° - ( x + y ) …(i)
Let ∠ABP = a and ∠ACP = b
then,
∠BAC = 180° - ( x + a ) - ( y + b )
⇒ ∠BAC = 180° - ( x + y ) - ( a + b )
⇒ ∠BAC = ∠BPC - ( a + b )
⇒ ∠BPC = ∠BAC + ( a + b )
⇒ ∠BPC > ∠BAC.
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