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

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by nadi

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by nadi

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^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.