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In a triangle ABC, the internal bisectors of angles B and C meet at P and the external bisectors of angles B and C meet at Q. Prove that : angle BPC + ANGLE BQC = 2 RIGHT ANGLES.
?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)


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