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In 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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In 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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It is given that PBC < QCB

We know that ABC + PBC = 180°

So, ABC = 180° - PBC

ACB + QCB = 180°

Therefore ACB = 180° -QCB

Now, PBC < QCB,

∴ ABC > ACB

Hence, AC > AB as sides opposite to the larger angle is always larger.

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