In sides AB and AC of ∆ABC are extended to points P and Q respective

1.	In sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Answer» Data : AB and AC are the sides of ∆ABC and AB and AC are produced to P and Q respectively. ∠PBC < ∠QCB. To Prove: AC > AB Proof: ∠PBC < ∠QCB Now, ∠PBC + ∠ABC = 180° ∠ABC = 180 – ∠PBC ………. (i) Similarly, ∠QCB + ∠ACB = 180° ∠ACB = 180 – ∠QCB …………. (ii) But, ∠PBC < ∠QCB (Data) ∴ ∠ABC > ∠ACB Comparing (i) and (ii), AC > AB (∵ Angle opposite to larger side is larger)

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

Answer»

Data : AB and AC are the sides of ∆ABC and AB and AC are produced to P and Q respectively.

∠PBC < ∠QCB.

To Prove: AC > AB

Proof: ∠PBC < ∠QCB

Now, ∠PBC + ∠ABC = 180°

∠ABC = 180 – ∠PBC ………. (i)

Similarly, ∠QCB + ∠ACB = 180°

∠ACB = 180 – ∠QCB …………. (ii)

But, ∠PBC < ∠QCB (Data)

∴ ∠ABC > ∠ACB Comparing (i) and (ii),

AC > AB (∵ Angle opposite to larger side is larger)