Answer:
[tex]m\angle B=54^{\circ}[/tex]
[tex]m\angle BAD=36^{\circ}[/tex]
[tex]m\angle CDA=90^{\circ}[/tex]
[tex]BAC=72^{\circ}[/tex]
Step-by-step explanation:
Given:
[tex]\overline{AB}\cong \overline{AC},[/tex]
[tex]\overline{AD}[/tex] bisects angle BAC
[tex]m\angle C=54^{\circ}[/tex]
Triangle ABC is isosceles triangle, because [tex]\overline{AB}\cong \overline{AC}.[/tex] Angles adjacent to the base of isosceles triangle ABC are congruent.
Hence,
[tex]m\angle C=m\angle B=54^{\circ}[/tex]
The sum of the measures of all interior angles is 180°, so,
[tex]m\angle B+m\angle C+m\angle BAC=180^{\circ}\\ \\m\angle BAC=180^{\circ}-2\cdot 54^{\circ}=72^{\circ}[/tex]
Since [tex]\overline{AD}[/tex] bisects angle BAC, angles BAD and CAD are congruent by definition of angle bisector. So,
[tex]m\angle BAD=m\angle CAD=\dfrac{1}{2}m\angle BAC=\dfrac{1}{2}\cdot 72^{\circ}=36^{\circ}[/tex]
AD ia angle bisector in isosceles triangle drawn to the base, so it is the height. Thus, AD and BC are perpendicular. So,
[tex]m\angle CDA=90^{\circ}[/tex]
