ABC is a triangle. The bisector of the exterior angle at B and the bisector of angle C intersect each other at D.Prove angle BDC =1/2 angle A.
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solution :-
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)...
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)...
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