Answer
By induction (or by the diagonal method), any convex \(n\)-gon has interior angles summing to \(180(n-2)\) degrees for all \(n\ge3\). The inductive reasoning (in a more streamlined version) is often summarized as follows:
1. **Base Case (\(n=3\)):** A triangle’s angles sum to \(180\).
2. **Inductive Step (\(n\to n+1\)):** Take a convex \((n+1)\)-gon, pick a vertex, and draw a diagonal to form an \(n\)-gon plus one triangle. By the inductive hypothesis, the \(n\)-gon’s angles sum to \(180(n-2)\). Adding the triangle’s 180 degrees gives a total of \(180(n-2) + 180 = 180(n-1)\), which matches \(180\bigl((n+1)-2\bigr)\).
Thus, for all \(n>3\), a convex \(n\)-gon’s interior angles sum to \(180(n-2)\) degrees.
Work Step by Step
Below is a standard induction proof that the sum of the interior angles of any \(n\)-sided convex polygon is \(180(n - 2)\) degrees, for all integers \(n > 3\).
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## Statement
**Claim.** For any integer \(n > 3\), if \(P\) is a convex \(n\)-sided polygon, then the sum of its interior angles is \(180(n - 2)\) degrees.
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## Proof by Induction
### Base Case (\(n = 4\))
For \(n=4\), \(P\) is a convex quadrilateral. One can draw a diagonal between two non‐adjacent vertices, dividing the quadrilateral into two triangles. Since each triangle has interior angles summing to \(180\) degrees, the total is \(180 + 180 = 360\) degrees. This matches \(180(4 - 2) = 360\) degrees, so the base case holds.
### Inductive Step
Assume the statement is true for some integer \(n \ge 4\). That is, **inductive hypothesis**:
> “In any convex \(n\)-sided polygon, the sum of its interior angles is \(180(n - 2)\) degrees.”
We must prove it for \(n+1\). Let \(Q\) be a convex \((n+1)\)-sided polygon. Pick any vertex \(V\) of \(Q\), and draw a diagonal from \(V\) to another non‐adjacent vertex so as to split \(Q\) into two smaller convex polygons: one with \(k\) sides (\(3 \le k \le n\)) and the other with \((n+1) - (k-1)\) sides. (The diagonal connects one vertex to another, effectively “adding” one side to each sub‐polygon along the diagonal.)
Concretely, we might choose a vertex \(V\) and connect it to a non‐adjacent vertex \(W\). This diagonal partitions \(Q\) into two convex polygons:
1. \(P_1\) with \(k\) sides,
2. \(P_2\) with \((n+1) - (k - 1) = n+2 - k\) sides.
Because \(3 \le k \le n\), both polygons \(P_1\) and \(P_2\) have **fewer** than \((n+1)\) sides but at least 3 sides each. Hence each polygon satisfies the inductive hypothesis. By the inductive hypothesis:
- The interior angles of \(P_1\) sum to \(180(k - 2)\) degrees.
- The interior angles of \(P_2\) sum to \(180\bigl((n+2-k) - 2\bigr) = 180(n - k)\) degrees.
Add these two sums together, and note that we have effectively counted all interior angles of \(Q\) exactly once. (The diagonal is interior to \(Q\), so it doesn’t introduce any additional “exterior” angle.)
Thus, the sum of interior angles of \(Q\) is
\[
180(k - 2) + 180(n - k)
\;=\;
180\bigl(k - 2 + n - k\bigr)
\;=\;
180(n - 2).
\]
Therefore, for \((n+1)\)-sided convex polygon \(Q\), the sum of its interior angles is \(180(n - 1)\). Wait, we need to be precise: the formula is \(180\bigl((n+1)-2\bigr) = 180(n - 1)\). Let’s confirm:
\[
180(k - 2) + 180(n - k) = 180(n - 2).
\]
But we want \(180((n+1) - 2) = 180(n - 1)\). Are these equal? Notice that
\[
n - 2 \;=\; (n+1) - 3.
\]
There might be a slight mismatch in the indexing if we are not careful. Let’s clarify carefully:
- \(Q\) has \((n+1)\) sides.
- We want to show its interior angles sum to \(180\bigl((n+1) - 2\bigr) = 180(n - 1)\).
In the split:
- \(P_1\) has \(k\) sides, sum of angles = \(180(k - 2)\).
- \(P_2\) has \((n+1) - (k - 1)\) sides, sum of angles = \(180\Bigl(\bigl[(n+1) - (k-1)\bigr] - 2\Bigr)\).
Compute \((n+1) - (k-1) = n+1-k+1 = n+2-k\). Thus:
\[
\text{sum}(P_2) = 180\bigl((n+2-k) - 2\bigr) = 180(n - k).
\]
So the total:
\[
\text{sum}(Q) = 180(k - 2) + 180(n - k) = 180(k - 2 + n - k) = 180(n - 2).
\]
But we wanted \(180\bigl((n+1)-2\bigr) = 180(n - 1)\). Let’s see if \(180(n - 2)\) and \(180(n - 1)\) differ. Indeed, \(180(n - 2) = 180n - 360\) and \(180(n - 1) = 180n - 180\). There’s a difference of 180. The resolution is that the diagonal we drew forms **one extra triangle** inside \(Q\). Actually, the diagonal in an \((n+1)\)-gon partitions it into **two** polygons whose total number of sides is \((n+1) + 2\) if you count the diagonal as a new side for each sub‐polygon. So each sub‐polygon has fewer than \((n+1)\) sides, but we have to be precise about how many:
- \(P_1\) has \(k+1\) vertices (the diagonal is an extra side), but effectively it is a polygon of \(k\) sides in the sense of the original boundary plus the diagonal.
- \(P_2\) has \((n+1 - k) + 1\) vertices = \(n+2 - k\) sides in the same sense.
Hence by the inductive hypothesis, if each sub‐polygon truly has **strictly fewer** than \((n+1)\) sides, we can apply the formula to each and then add them.
**A Simpler Approach**: Instead of the “two sub-polygons” approach, do the more common approach:
1. Take a convex \((n+1)\)-gon.
2. Choose one vertex \(V\). Draw **all** diagonals from \(V\) to the other non‐adjacent vertices.
This splits the \((n+1)\)-gon into \(n-1\) triangles.
3. Each triangle has interior angles summing to 180, so the total is \((n-1)\times 180 = 180n - 180\).
But that’s exactly \(180((n+1)-2)\).
Hence we get \(180(n - 1)\) for an \((n+1)\)-gon, which is \(180\bigl((n+1)-2\bigr)\). That’s the standard straightforward proof. It also works as an induction:
- **Base case:** \(n=3\) (triangle). Sum of interior angles is \(180\). Matches \(180(3-2) = 180\).
- **Inductive step:** Going from \(n\)-gon to \((n+1)\)-gon, pick a vertex and draw one diagonal to create a smaller polygon. Then apply the inductive hypothesis to the smaller polygon, and note each diagonal adds another triangle (180 degrees). Summation yields the result.
