Prove the Polygon​ Interior-Angle Sum Theorem.

Question

Given​- A polygon with n sides.
Prove​- The sum of the measures of the interior angles is ​(nminus​2)180degrees.
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Part 1
Draw a polygon with n​ sides, and let the number of sides be a​ number, n. Choose a point P in the interior of the polygon and join P to each vertex of the polygon.
1. Since the polygon has n​ sides,
▼
n minus 1
2 n
n
StartFraction n Over 2 EndFraction
triangles are formed.

Answer 1

To prove the Polygon Interior-Angle Sum Theorem, we start with a polygon that has \( n \) sides and follow the steps you mentioned.

Step 1: Draw the Polygon

Let’s denote the polygon as \( ABCD... \) up to \( n \) vertices. We have \( n \) sides and \( n \) vertices.

Step 2: Connect a Point to Each Vertex

Choose a point \( P \) inside the polygon and connect it to each vertex of the polygon by drawing line segments \( PA, PB, PC, \ldots, P_n \).

Step 3: Number of Triangles Formed

Because we have connected \( P \) to each of the \( n \) vertices, we form \( n \) triangles. Specifically, these triangles are:

• \( \triangle PAB \)
• \( \triangle PBC \)
• \( \triangle PCD \)
• ...
• \( \triangle PA_nB_1 \)

Thus, the polygon with \( n \) sides forms \( n \) triangles.

Step 4: Interior Angles of Each Triangle

The sum of the interior angles of each triangle is \( 180^\circ \). Since there are \( n \) triangles, the total sum of the interior angles of these triangles can be calculated as follows:

\[ \text{Sum of interior angles of all triangles} = n \times 180^\circ = 180n \]

Step 5: Relate Triangle Angles to Polygon Angles

When we consider the interior angles of the polygon, we see that each interior angle of the polygon (let's denote the angles as \( A_1, A_2, ..., A_n \)) is formed by adjacent pairs of angles in the triangles. There are \( n \) angles in the polygon, but the \( P \) angles at point \( P \) are included in the total sum of angles of the triangles.

Step 6: Counting Angle Contributions

When we consider the sum of the angles:

• Each interior angle of the polygon contributes to the sum, but we also have \( P \) contributing angles.
• For \( n \) triangles, the \( P \) angles combine to constitute the angles around \( P \).

Thus, the angles at point \( P \) form a full circle. Therefore, the total contribution from all angles at point \( P \) is \( 360^\circ \).

Final Calculation

Subtracting the total angle contribution at \( P \) from the total angles we obtained from the triangles gives us:

\[ \text{Sum of interior angles of the polygon} = \text{Sum of angles from triangles} - \text{Angles at P} \] \[ = 180n - 360 \] \[ = 180(n - 2) \]

Conclusion

Therefore, we conclude that the sum of the measures of the interior angles of a polygon with \( n \) sides is:

\[ (n - 2) \times 180^\circ \]

This completes the proof of the Polygon Interior-Angle Sum Theorem.