Question

In a triangle ABC, the internal bisectors of angles Band C meet at P and the external bisectors of the angles Band C meet at Q. Prove that : angle BPC+ angle BQC=180°

A detailed explanation would do just awesome!

Answer 1

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∠ACB and ∠QCE form a linear pair



Hence ∠ACB+∠QCB =180º

⇒1/2∠ACB + 1/2∠QCB = 90°

⇒ ∠PCB+∠BCQ = 90° ( As PC and QC are angle bisectors)

⇒ ∠PCQ = 90°

Similarly it can be proven that

∠PBQ = 90°



now in quadrilateral BPCQ, sum of all the four angles = 360º

⇒ ∠BPC+∠PCQ+∠CQB+∠QBP=360º

⇒ ∠BPC+∠BQC = 180º ( as ∠PCQ=∠PBQ=90º)