Question

given △abc with angle bisectors \\(\overline{ae}\\), \\(\overline{bf}\\), and \\(\overline{cd}\\), and incenter point o.
if \\( m\angle a = 100^\circ \\) and \\( m\angle abf = 30^\circ \\), determine \\( m\angle acd \\).
\\( m\angle acd = \square ^\circ \\)

Explanation:

Step1: Find \(m\angle ABC\)

Since \(BF\) is an angle - bisector of \(\angle ABC\), if \(m\angle ABF = 30^{\circ}\), then \(m\angle ABC=2\times m\angle ABF\).
\(m\angle ABC = 60^{\circ}\)

Step2: Use the angle - sum property of a triangle

The sum of the interior angles of a triangle \(\triangle ABC\) is \(180^{\circ}\). We know that \(m\angle A + m\angle ABC+m\angle ACB=180^{\circ}\). Given \(m\angle A = 100^{\circ}\) and \(m\angle ABC = 60^{\circ}\).
Substitute the values into the formula: \(100^{\circ}+60^{\circ}+m\angle ACB = 180^{\circ}\).
Solve for \(m\angle ACB\): \(m\angle ACB=180^{\circ}-(100^{\circ} + 60^{\circ})=20^{\circ}\)

Step3: Find \(m\angle ACD\)

Since \(CD\) is an angle - bisector of \(\angle ACB\), then \(m\angle ACD=\frac{1}{2}m\angle ACB\)
\(m\angle ACD=\frac{1}{2}\times20^{\circ}=10^{\circ}\)

Answer:

\(10\)