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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) |