Question

given △abc with angle bisectors \\(\overline{ae}\\), \\(\overline{bf}\\), and \\(\overline{cd}\\), and incenter o.

triangle image

if \\( m\angle a = 74^\circ \\) and \\( m\angle abf = 27^\circ \\), determine \\( m\angle acd \\).

if the \\( m\angle acd = 54^\circ \\), then go to question 15.

if the \\( m\angle acd = 52^\circ \\), then go to question 2.

if the \\( m\angle acd = 74^\circ \\), then go to question 4.

if the \\( m\angle acd = 26^\circ \\), then go to question 3.

Explanation:

Step1: Recall triangle angle sum and angle bisectors

The sum of angles in a triangle is \( 180^\circ \). Angle bisectors divide angles into two equal parts. \( \overline{BF} \) is an angle bisector, so \( m\angle ABC = 2 \times m\angle ABF \).

Step2: Calculate \( m\angle ABC \)

Given \( m\angle ABF = 27^\circ \), then \( m\angle ABC = 2\times27^\circ = 54^\circ \).

Step3: Calculate \( m\angle ACB \)

We know \( m\angle A = 74^\circ \) and \( m\angle ABC = 54^\circ \). Using the triangle angle sum formula \( m\angle A + m\angle ABC + m\angle ACB = 180^\circ \), we solve for \( m\angle ACB \):

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Step4: Determine \( m\angle ACD \)

Since \( \overline{CD} \) is an angle bisector, it divides \( \angle ACB \) into two equal angles. Wait, no—wait, earlier calculation might have an error. Wait, let's re - check. Wait, \( \angle A = 74^\circ \), \( \angle ABF = 27^\circ \), so \( \angle ABC = 2\times27 = 54^\circ \). Then \( \angle ACB=180 - 74 - 54 = 52^\circ \)? No, wait, no—wait, the incenter, so \( CD \) bisects \( \angle ACB \)? Wait, no, wait the problem is to find \( m\angle ACD \). Wait, maybe I made a mistake. Wait, let's start over.

The sum of angles in a triangle: \( \angle A+\angle B+\angle C = 180^\circ \). \( BF \) is an angle bisector, so \( \angle ABF=\angle FBC = 27^\circ \), so \( \angle B = 2\times27 = 54^\circ \). \( \angle A = 74^\circ \), so \( \angle C=180 - 74 - 54 = 52^\circ \)? No, that's not right. Wait, no, the question is to find \( \angle ACD \). Wait, \( CD \) is an angle bisector of \( \angle C \)? Wait, no, maybe \( CD \) bisects \( \angle C \), so \( \angle ACD=\frac{1}{2}\angle C \). Wait, but let's recalculate \( \angle C \).

Wait, \( \angle A = 74^\circ \), \( \angle ABF = 27^\circ \), so \( \angle ABC = 2\times27 = 54^\circ \). Then \( \angle ACB=180 - 74 - 54 = 52^\circ \). But if \( CD \) is an angle bisector, then \( \angle ACD=\frac{1}{2}\angle ACB \)? No, that would be \( 26^\circ \). Wait, maybe I messed up the angle bisector. Wait, the incenter is the intersection of angle bisectors. So \( AE \) bisects \( \angle A \), \( BF \) bisects \( \angle B \), \( CD \) bisects \( \angle C \).

So \( \angle A = 74^\circ \), so \( \angle BAE=\angle EAC = 37^\circ \). \( \angle ABF = 27^\circ \), so \( \angle FBC = 27^\circ \), so \( \angle B = 54^\circ \). Then \( \angle C=180 - 74 - 54 = 52^\circ \). Then \( CD \) bisects \( \angle C \), so \( \angle ACD=\frac{1}{2}\angle C=\frac{52^\circ}{2}=26^\circ \). Ah, there we go. So \( m\angle ACD = 26^\circ \).

Answer:

\( 26^\circ \)