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

`m\angle B=54^(\circ)`

`m\angle BAD=36^(\circ)`

`m\angle CDA=90^(\circ)`

`BAC=72^(\circ)`

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=(1)/(2)m\angle BAC=(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)`