Question

1. in the accompanying figure, \\( \overline { a b } \cong \overline { b c } \\), \\( m \angle a = 40 \\), and \\( \overline { c d } \\) bisects \\( \angle a c b \\). find \\( m \angle c d b \\).

Explanation:

Step1: Find the measure of \(\angle ACB\) in \(\triangle ABC\)

Since \(\overline{AB}\cong\overline{BC}\), \(\triangle ABC\) is isosceles with \(\angle A=\angle ACB = 40^{\circ}\) (base angles of an isosceles triangle are equal).

Step2: Find the measure of \(\angle DCB\)

Given that \(\overline{CD}\) bisects \(\angle ACB\). By the definition of an angle - bisector, \(\angle DCB=\frac{1}{2}\angle ACB\). Substituting \(\angle ACB = 40^{\circ}\), we get \(\angle DCB=\frac{1}{2}\times40^{\circ}=20^{\circ}\).

Step3: Find the measure of \(\angle B\) in \(\triangle ABC\)

Using the triangle - angle sum theorem (\(\angle A+\angle B+\angle ACB = 180^{\circ}\)). Substituting \(\angle A = 40^{\circ}\) and \(\angle ACB = 40^{\circ}\), we have \(40^{\circ}+\angle B + 40^{\circ}=180^{\circ}\). Then \(\angle B=180^{\circ}-(40^{\circ}+40^{\circ}) = 100^{\circ}\).

Step4: Find the measure of \(\angle CDB\) in \(\triangle BCD\)

Using the triangle - angle sum theorem in \(\triangle BCD\) (\(\angle B+\angle DCB+\angle CDB = 180^{\circ}\)). Substituting \(\angle B = 100^{\circ}\) and \(\angle DCB = 20^{\circ}\), we get \(100^{\circ}+20^{\circ}+\angle CDB=180^{\circ}\). Then \(\angle CDB=180^{\circ}-(100^{\circ}+20^{\circ})=60^{\circ}\).

Answer:

\(60^{\circ}\)