Question

In triangle ABC, angleABC> angleACB. sides AB andAC are extended to points P and Q respectively. Prove that angle PBC < angleQCB.

Answer 1

here,
^ABC+∨PBC=180° (Property of a Line)
&
∧QCB+∨ACB=180°
On subtracting above equations,
∧ABC-∨ACB=∧QCB-∨PBC
We Know that ∧ABC>∨ACB⇒∧ABC-∨ACB>0
⇒∧QCB-∨PBC>0⇒∧QCB>∨PBC which is same as ∨PBC<∧QCB.


here another thing which proves that PBC is smaller is that QCB is Obtuse and PBC is acute.Diagram makes it clear too.However proof is required as it is asked.