Question

plz answer the question it is very important....

Answer 1

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