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Question
In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
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Solution
In the given figure,
∠ABC + ∠PBC = 180° (Linear pair)
⇒ ∠ABC = 180° − ∠PBC ... (1)
Also,
∠ACB + ∠QCB = 180°
∠ACB = 180° − ∠QCB … (2)
As ∠PBC < ∠QCB,
⇒ 180º − ∠PBC > 180º − ∠QCB
⇒ ∠ABC > ∠ACB [From equations (1) and (2)]
⇒ AC > AB (Side opposite to the larger angle is larger.)
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