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Let PQR be any traingle and PM be the bisector of vertical angle QPR ( see image )
Now,
PQ = PR .... (isosceles traingle
Angle QPM = Angle RPM ... (Given)
PM = PM ... (common side)
Therefore triangle QPM = triangle RPM ... (by SAS test of similarity )
Therfore angle QMP= angle RMP = 90degree .. (angle QMP + angle RMP = 180)
Therefore PM bisects the base at right angle
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