BD is the angle bisector of

Note: Let us consider, we need to find the [tex]m\angle ABC[/tex] and [tex]m\angle DBC[/tex].
Given:
In the given figure, BD is the angle bisector of ABC.
To find:
The [tex]m\angle ABC[/tex] and [tex]m\angle DBC[/tex].
Solution:
BD is the angle bisector of ABC. So,
[tex]m\angle ABD=m\angle DBC[/tex]
[tex]3x=x+20[/tex]
[tex]3x-x=20[/tex]
[tex]2x=20[/tex]
Divide both sides by 2.
[tex]x=\dfrac{20}{2}[/tex]
[tex]x=10[/tex]
Now,
[tex]m\angle DBC=(x+20)^\circ[/tex]
[tex]m\angle DBC=(10+20)^\circ[/tex]
[tex]m\angle DBC=30^\circ[/tex]
And,
[tex]m\angle ABC=(3x)^\circ+(x+20)^\circ[/tex]
[tex]m\angle ABC=(4x+20)^\circ[/tex]
[tex]m\angle ABC=(4(10)+20)^\circ[/tex]
[tex]m\angle ABC=(40+20)^\circ[/tex]
[tex]m\angle ABC=60^\circ[/tex]
Therefore, [tex]m\angle DBC=30^\circ,m\angle ABD=30^\circ[/tex] and [tex]m\angle ABC=60^\circ[/tex].