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← Polygons and Angles

Rule

Polygon Exterior Angles Theorem

The sum of the measures of the exterior angles of a polygon, one angle at each vertex, is

36 0^{\circ} .

Hexagon with all the exterior angles marked

Based on the diagram, the relation below holds true.

$m ∠ 1 +$ $m ∠ 2 +$ $m ∠ 3 +$ $m ∠ 4 +$ $m ∠ 5 +$ $m ∠ 6 = 36 0^{\circ}$

Proof

Convex Polygon

Consider a hexagon and all its exterior angles. Let $S$ be the sum of the measures of the exterior angles.

Notice that an interior angle and its exterior angle form a linear pair. Therefore, the sum of their measures is equal to

18 0^{\circ} .

⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ m ∠ 1 + m ∠ A = 18 0^{\circ} m ∠ 2 + m ∠ B = 18 0^{\circ} m ∠ 3 + m ∠ C = 18 0^{\circ} m ∠ 4 + m ∠ D = 18 0^{\circ} m ∠ 5 + m ∠ E = 18 0^{\circ} m ∠ 6 + m ∠ F = 18 0^{\circ}

By the Polygon Interior Angles Theorem, the sum of the measures of the interior angles of a polygon with

n

sides is

(n - 2) \cdot 18 0^{\circ} .

With this information, add all the equations in the system above.

m ∠ 1 + m ∠ A m ∠ 2 + m ∠ B m ∠ 3 + m ∠ C m ∠ 4 + m ∠ D m ∠ 5 + m ∠ E + m ∠ 6 + m ∠ F S + (6 - 2) \cdot 18 0^{\circ} = 18 0^{\circ} = 18 0^{\circ} = 18 0^{\circ} = 18 0^{\circ} = 18 0^{\circ} = 18 0^{\circ} = 6 \cdot 18 0^{\circ}

Finally, the last equation can be solved for

S .

S + (6 - 2) \cdot 18 0^{\circ} = 6 \cdot 18 0^{\circ}

▼

Solve for

S

Subtract terms

S + 4 \cdot 18 0^{\circ} = 6 \cdot 18 0^{\circ}

Multiply

S + 72 0^{\circ} = 108 0^{\circ}

$LHS - 72 0^{\circ} = RHS - 72 0^{\circ}$

S = 36 0^{\circ}

The desired result is obtained.

$S = 36 0^{\circ}$

Although the proof above used a hexagon, the same reasoning can be applied no matter the number of sides of the polygon.

Proof

Concave Polygon

Consider a concave polygon and one exterior angle at each vertex. Let $S$ be the sum of the measures of the exterior angles.

Notice that

∠ 3

lies inside the polygon. In this case, its measure is considered to be negative. Because of this,

∠ 3

and

∠ C

are supplementary.

m ∠ C + m ∠ 3 = 18 0^{\circ} (I)

The rest of exterior angles form a linear pair with their corresponding interior angles. This means all those pair of angles are supplementary.

⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ m ∠ 1 + m ∠ A = 18 0^{\circ} m ∠ 2 + m ∠ B = 18 0^{\circ} m ∠ 4 + m ∠ D = 18 0^{\circ} m ∠ 5 + m ∠ E = 18 0^{\circ}

Next, add these four equations to Equation (I). Remember that the Polygon Interior Angles Theorem states that the sum of the measures of the interior angles of a polygon with

n

sides is

(n - 2) \cdot 18 0^{\circ} .

m ∠ 1 + m ∠ A m ∠ 2 + m ∠ B m ∠ 3 + m ∠ C m ∠ 4 + m ∠ D + m ∠ 5 + m ∠ E S + (5 - 2) \cdot 18 0^{\circ} = 18 0^{\circ} = 18 0^{\circ} = 18 0^{\circ} = 18 0^{\circ} = 18 0^{\circ} = 5 \cdot 18 0^{\circ}

Finally, solve the resulting equation for

S .

S + (5 - 2) \cdot 18 0^{\circ} = 5 \cdot 18 0^{\circ}

▼

Solve for

S

Subtract terms

S + 3 \cdot 18 0^{\circ} = 5 \cdot 18 0^{\circ}

Multiply

S + 54 0^{\circ} = 90 0^{\circ}

$LHS - 54 0^{\circ} = RHS - 54 0^{\circ}$

S = 36 0^{\circ}

As can be seen, the sum of the measures of the exterior angles of the concave polygon is

36 0^{\circ},

which completes the proof. Keep in mind that the same reasoning can be applied to any other concave polygon.

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