In Fig. 7.48, sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB.

It is given in the question that:

PBC < QCB


Now,


ABC + PBC = 180o


ABC = 180o - PBC


Also,


ACB + QCB = 180o


ACB = 180o - QCB


Since,


PBC < QCB therefore,


ABC > ACB


Thus,


AC > AB as sides opposite to the larger angle is larger


6