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12/07/2021 12:17 pm
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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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12/07/2021 12:18 pm
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.