How to prove the inner angles of an n sided and the formula n 2 180

From any vertex to non-adjacent vertices, the n-sided can get (n-2) triangles, and the sum of the inner angles of all triangles is the sum of the internal angles of this polygon, and the sum of the internal angles of the triangle is 180, so the sum of the inner angles of the n-sided triangle is 180°.

Method 2: Select any point inside and connect all vertices to get n triangles, and the sum of the inner angles of the polygon = n the inner angles of the triangle and -360 (that is, the selected point is the sum of all the angles of the vertices) = (n-2) 180

Method 1: As shown in Figure D27-1-2, take any point O in the n-sided shape, connect the line segments of O and each vertex, and divide the n-sided shape into n triangles. Since the sum of the internal angles of the n triangles is equal to n·180°, and the sum of the n angles with o as the common vertex is 360°, the sum of the internal angles of the n-sided is n·180°-2 180°=(n-2)·180°

The sum of the internal angles of the n-sided shape is equal to (n-2) 180°

Evidence 2: As shown in Figure D27-1-3, through any vertex A1 of the polygon, connect the line segments of point A1 and each vertex, and divide the N-sided into (N-2) triangles. Since the sum of the internal angles of the (n-2) triangle is equal to (n-2)·180°, the sum of the internal angles of the n-sided is (n-2) 180°

Method 3: As shown in Figure D27-1-4, take any point p on the side of the n-sided A1A2, and the line segment connecting the p point and each vertex can divide the n-sided into (n-1) triangles, and the sum of the inner angles of the (n-1) triangle is equal to (n-1)·180°The sum of the angles of (n-1) with p as the common vertex is 180°, so the sum of the inner angles of the n-sided is (n-1)·180°-180°=(n-2)·180°

From any vertex to a non-adjacent vertice, the n-sided can get (n-2) triangles megarent, and the sum of the inner angles of all the alternate triangles is the sum of the inner angles of this polygon, and the sum of the inner angles of the triangle is 180, so the sum of the inner angles of the n-sided triangle is 180°Method 2: Select any point inside and connect all vertices to ,..

1,2 sides are not counted, the sum of the inner angles of the triangle is 180 degrees (this proof is very simple), and then any polygon can be divided into (n-2) triangles.

So to get the above conclusion.

can only be so dissolved and released.

As for why it can be separated.

You can find any point in the polygon, and then exclude two adjacent points from Lu Shihe.

Then use this point to connect with other points.

You can connect (n-3) lines.

Make (n-2) triangles.

There are three ways to prove it:

1. Starting from the inner point o of the n-sided to connect the vertices of the n-sided to obtain n triangles, the sum of the internal disadvantages of the rent-withered angles is n*180°, and the sum of the n angles at the subtract o is 360°, and the sum of the inner angles of the n-sided is obtained: n*180°-360°=(n-2)*180°;

2. Starting from a vertex of the n-sided shape, connect the diagonals (n-3) to obtain (n-2) three-lease hole angles, and the sum of the inner angles of the (n-2) triangle is the sum of the inner angles of the n-side, that is, the sum of the inner angles of the n-sided shape is: (n-2)*180°;

3. The sedan car takes a point from one side of the n-sided shape, connects the other vertices (n-2) to obtain (n-1) triangles, and subtracts a flat angle from the sum of the inner angles of the (n-1) triangle to obtain the sum of the inner angles of the n-side.

n-1)*180°-180°=(n-2)*180°.

The n-sided shape takes a vertex as the fixed point, and can connect the Min trap line to n-3 points (except for the point and the adjacent point) to obtain n-2 posture height triangles, so the sum of the inner angles is 180 degrees and multiplied by n-2

Take a point O from the macro inside the polygon and connect the O to the vertices. Each vertex and O point form a number of triangles. N vertices form n triangles.

The sum of the inner angles of the triangle is 180n, the sum of the inner angles of the triangle is n*180, and the sum of the vertices of the n triangles around the point of o is 360, so the sum of the inner angles of the polygon is n*180-360=(n-2)*180

The sum of the outer angles of the n-sided shape is equal to 360 degrees, divided by n, is an outer angle of 360 n, an inner angle is 180-360 n, multiplied by n, it is n(180-360 n) = 180n-360, extract 180, and obtain (n-2)·180°

Any n-sided has n vertices, starting from one of them, you can connect n-3 diagonals to divide the n-2 triangles, and the sum of the inner angles of each triangle is 180°, so the sum of the inner angles of the n-sided is (n-2) 180°

Starting from one of the vertices of the n-side, we can draw (n-3) diagonals, which divide the n-side into (n-2) triangles, and the sum of the inner angles of a triangle is 180 degrees, so the sum of the inner corners of the n-sided chain is equal to 180*(n-2).