Question

in the given figure point D is in the interior of ∆ABC, Angle A=70°. BD & CD are the bisectors of angle ABC & ACB respectively find angle BDC and prove that BDC>A​

Answer 1

Answer:

hey mate your answer with image

Step-by-step explanation:

Extend the line BC to E

BD and CD are angular bisectors,

∴∠ABD=∠DBC=x and ∠ACD=∠DCE=y

∠ABC=2x and ∠ACE=2y

Consider △ABC,

∠ACE=∠ABC+∠BAC ------exterior angle is equal to sum of interior opposite angle

2y=2x+∠A

y−x=2∠A------(i)

Consider △BCD,

∠DCE=∠DBC+∠BDC ------exterior angle is equal to sum of interior opposite angle

y=x+∠D

y−x=∠D------(ii)

From(i) and (ii)

∠D=21 ∠A

Option A. Rest are not true.

hope it helps you