Question

Please anybody help me please this is so hard that not even I can't solve this I don't pay attention in class

Answer 1

Answer:

$m\angle B=54^{\circ}$

$m\angle BAD=36^{\circ}$

$m\angle CDA=90^{\circ}$

$BAC=72^{\circ}$

Step-by-step explanation:

Given:

$\overline{AB}\cong \overline{AC},$

$\overline{AD}$ bisects angle BAC

$m\angle C=54^{\circ}$

Triangle ABC is isosceles triangle, because $\overline{AB}\cong \overline{AC}.$ Angles adjacent to the base of isosceles triangle ABC are congruent.

Hence,

$m\angle C=m\angle B=54^{\circ}$

The sum of the measures of all interior angles is 180°, so,

$m\angle B+m\angle C+m\angle BAC=180^{\circ}\\ \\m\angle BAC=180^{\circ}-2\cdot 54^{\circ}=72^{\circ}$

Since $\overline{AD}$ bisects angle BAC, angles BAD and CAD are congruent by definition of angle bisector. So,

$m\angle BAD=m\angle CAD=\dfrac{1}{2}m\angle BAC=\dfrac{1}{2}\cdot 72^{\circ}=36^{\circ}$

AD ia angle bisector in isosceles triangle drawn to the base, so it is the height. Thus, AD and BC are perpendicular. So,

$m\angle CDA=90^{\circ}$