$n$ points on a circle connect by lines, the sum of the internal angles is $(n - 1) 180$°

Question

I am trying to understand a proof by induction on a geometry problem:

For all $n\geq 3$, if $n$ distinct points on a circle are connected in consecutive order by straight lines, then the interior angles of the resulting polygon (n-gon) add up to $(n - 2) 180$°.

Base case:

Let $n = 3$

It is well known that the interior angles of a triangle add up to 180°.

Induction hypothesis:

Assume the statement is true for $n$ points: $(A(n) = (n-2)180$°$)$

Induction step:

Prove the statement for $(A(n + 1) = (n - 1)180$°$)$

[$(n - 1)180$° is the same as $(n - 1) \cdot 180$°, right?]

This is the image that is given:

The proof is the following:

• $P_{n+1}$ polygon formed by $n+1$ points on the circle. [ok]

• $P_n$ the polygon derived from $P_{n+1}$ by skipping the last point [ok]

• By induction hypothesis $A(n) = (n-2)180$° [ok]

• $P_{n+1}$ = $P_n$ + triangle$(A_1, A_n, A_{n+1})$ [not so ok the triangle part]

• Thus, $A(n+1)$ = $(n-2)180 + 180 = (n-1)180$ [not so ok]

The parts [not so ok] are the parts I am not understanding well. Could you please explain them in another way?

Answer 1

When you go from polygon $P_{n}$ to polygon $P_{n+1}$, you add remove two angles and add up three.

You remove $A_{n-1}A_nA_1$ and $A_nA_1A_2$ and add $A_{n-1}A_nA_{n+1}$, $A_nA_{n+1}A_1$ and $A_{n+1}A_nA_2$.

It is easy to see that the sum of the three new angles is equal to the sum of the two previous one plus the sum of the internal angles of the triangle $A_nA_{n+1}A_1$.

Answer 2

If you look carefully at the picture you will notice that the result you need is the sum of the angles of the polygon $P_n + \angle{A_1A_nA_{n+1}} + \angle{A_1A_{n+1}A_{n}} + \angle{A_nA_1A_{n+1}}$, which are the three angles of the triangle $A_1A_{n}A_{n+1}$ and therefore they sum up to $180^°$.