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Question
In ΔABC, the exterior ∠PBC > exterior ∠QCB. Prove that AB > AC.
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Solution
It is given that ∠PBC > ∠QCB -------(1)
∠PBC + ∠ABC = 180 ...[Linear pair angles]
⇒ ∠PBC = 180 - ∠ABC
Similarly, ∠QCB = 180 - ∠ACB
From (1) and (2)
180 - ∠ABC > 180 - ∠ACB
⇒ -∠ABC > - ∠ACB
⇒ ABC < ∠ACB or ∠ACB > ∠ABC
It is known that in a triangle, the greater angle has the longer side opposite to it.
∴ AB > AC.
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