Question

Prove that the angle formed by the bisector of interior angle A and the bisector of exterior angle B of a triangle is half of angle C

PLEASE URGENT ANS DO IT CLEARLY ON A PAGE

Answer 1

Step-by-step explanation:

here, given :-

ABC is a triangle.

BD is the interior angle bisector of ∠B and CD is the exterior angle bisector of ∠C.

To prove that :- BDC =1/2 angle A.

proof :- In ΔABC

∠A + ∠B + ∠C = 180°

⇒ ∠A = 180° – x – y .... (1)

and In ΔBDE

angle B = x / 2

angle c= y + 180- y/2= 180+y/2

and , therefore angle b + c + d = 180°

so, angle D = 180°- x /2 - 180+y/2

= 180 - x - y /2 ..........( 2 )

Now from eqation ( 1 ) and ( 2 ), we get the result..

angle D = angle A / 2... (ANS)...

Answer 2

Step-by-step explanation:

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