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)
It is given that∠PBC<∠QCB ⇒180∘−∠PBC > 180∘−∠QCB ⇒∠ABC > ∠ACB [From (1) and (2)] ∴ AC>AB.(In any triangle, the side opposite to the larger angle is longer)