Given:
AD is an angle bisector in triangle ABC. [tex]m\angle CAB=44^\circ, m\angle ACB=72^\circ, m\angle ABC=64^\circ[/tex].
To find:
The value of [tex]m\angle ADC[/tex].
Solution:
AD is an angle bisector in triangle ABC.
[tex]m\angle CAD=m\angle BAD=\dfrac{m\angle CAB}{2}[/tex]
[tex]m\angle CAD=m\angle BAD=\dfrac{44^\circ}{2}[/tex]
[tex]m\angle CAD=m\angle BAD=22^\circ[/tex]
According to the angle sum property, the sum of all interior angles of a triangle is 180 degrees.
Using angle sum property in triangle CAD, we get
[tex]m\angle CAD+m\angle ADC+m\angle ACB=180^\circ[/tex]
[tex]22^\circ+m\angle ADC+72^\circ=180^\circ[/tex]
[tex]m\angle ADC+94^\circ=180^\circ[/tex]
[tex]m\angle ADC=180^\circ-94^\circ[/tex]
[tex]m\angle ADC=86^\circ[/tex]
Therefore, the angle of angle ADC is [tex]86^\circ[/tex].
