Question

please answer yaar m.......

Answer 1

In the Given figure,


angle ACD lies exterior to ∆ABC

By using external angle property, we can say,

angle BAC + angle ABC = angle ACD


Now given that EC is the bisector of angle ACD

=> 1/2 angle ACD = angle ECD

and BE is the bisector of angle ABC

=> 1/2angle ABC = angle EBC

Now since,

angle BAC + angle ABC = angle ACD

=> 1/2(angle BAC + angle ABC) = 1/2 angle ACD

=> 1/2angle BAC + 1/2 angle ABC = 1/2 angle ACD

=> 1/2 angle BAC + angle EBC = angle ECD

Again, in ∆EBC,

angle ECD lies exterior to ∆EBC, so using exterior angle property of triangle we can write,

angle EBC + angle BEC = angle ECD


But we derived above that

angle ECD = 1/2 angle BAC + angle EBC

so we can write,

1/2ang BAC + ang EBC = ang EBC +ang BEC

=> 1/2 ang BAC = ang EBC + ang BEC - angEBC

Now cancelling +ang EBC with - angEBC


we get,


1/2 angle BAC = angle BEC

Hence Proved


Hope it helps dear friend ☺️

Answer 2

hey
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hope helped
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