Question

given △abc with angle bisectors (overline{ae}), (overline{bf}), and (overline{cd}), and incenter point o. if ( mangle a = 74^circ ) and ( mangle abf = 27^circ ), determine ( mangle acd ). ( mangle acd = square^circ )

Explanation:

Step1: Find \(m\angle ABC\)

Since \(BF\) is an angle - bisector, \(m\angle ABC = 2m\angle ABF\).
Given \(m\angle ABF=27^{\circ}\), then \(m\angle ABC = 2\times27^{\circ}=54^{\circ}\).

Step2: Use the angle - sum property of a triangle

In \(\triangle ABC\), we know that \(m\angle A+m\angle ABC + m\angle ACB=180^{\circ}\).
Given \(m\angle A = 74^{\circ}\) and \(m\angle ABC = 54^{\circ}\).
Substitute into the formula: \(74^{\circ}+54^{\circ}+m\angle ACB = 180^{\circ}\).
Then \(m\angle ACB=180^{\circ}-(74^{\circ} + 54^{\circ})=180^{\circ}-128^{\circ}=52^{\circ}\).

Step3: Find \(m\angle ACD\)

Since \(CD\) is an angle - bisector of \(\angle ACB\), \(m\angle ACD=\frac{1}{2}m\angle ACB\).
Substitute \(m\angle ACB = 52^{\circ}\), then \(m\angle ACD=\frac{1}{2}\times52^{\circ}=26^{\circ}\).

Answer:

\(26\)