In Δ ABC, A=50° and BC is produced to a point D. If the bisectors of ABC and ACD meet at E, then E =

In Δ ABC

A + B + C = 180o


50o + B + C = 180o


B + C = 180o – 50o


B + C = 10o (i)


In


E + BCE + EBC = 180o


E + 180o – ( ACD) + B = 180o (ii)


By exterior angle theorem,


ACD = 50o + B


Putting value of ACD in (ii), we get


E + 180o - (50o + B) + B = 180o


E – 25o - B + B = 0


E – 25o = 0


E = 25o

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