Question

given abc is a triangle
prove m∠fbc + m∠icd + ∠cae = 360°.
you may use the format shown below. draw a table using the insert table tool from your submission dropbox menu.
statements
reasons

Explanation:

Step1: Recall Exterior Angle Sum

For any triangle, the sum of exterior angles is \( 360^\circ \). Let's identify the exterior angles: \( \angle FBC \), \( \angle ICD \), \( \angle CAE \) are exterior angles of \( \triangle ABC \).

Step2: Use Exterior Angle Property

Each exterior angle and its adjacent interior angle form a linear pair (sum to \( 180^\circ \)). Let interior angles be \( \angle ABC \), \( \angle ACB \), \( \angle BAC \). So:

• \( m\angle FBC + m\angle ABC = 180^\circ \) (Linear Pair)
• \( m\angle ICD + m\angle ACB = 180^\circ \) (Linear Pair)
• \( m\angle CAE + m\angle BAC = 180^\circ \) (Linear Pair)

Step3: Sum the Equations

Add the three equations:
\( (m\angle FBC + m\angle ABC) + (m\angle ICD + m\angle ACB) + (m\angle CAE + m\angle BAC) = 180^\circ + 180^\circ + 180^\circ \)
Simplify left side: \( m\angle FBC + m\angle ICD + m\angle CAE + (m\angle ABC + m\angle ACB + m\angle BAC) \)
Right side: \( 540^\circ \)

Step4: Use Triangle Angle Sum

In \( \triangle ABC \), \( m\angle ABC + m\angle ACB + m\angle BAC = 180^\circ \) (Triangle Angle Sum). Substitute:
\( m\angle FBC + m\angle ICD + m\angle CAE + 180^\circ = 540^\circ \)
Subtract \( 180^\circ \) from both sides:
\( m\angle FBC + m\angle ICD + m\angle CAE = 360^\circ \)

Answer:

The sum of the exterior angles \( m\angle FBC + m\angle ICD + m\angle CAE = 360^\circ \) is proven using linear pairs and triangle angle sum property.