The proof I did was not one of three I saw. I'll describe the proofs. Which would you use to teach students?
1. The angles from a through e are 36 degrees each because they are inscribed in 72 degree arcs. I don't like circles, so I would never work with them unless I had to.
2. The pentagon has five 108 degree angles, and each triangle has two angles supplementary to 108 degrees, meaning they are 72 degrees each, and angles a through e are 36 degrees each as part of 367272 triangles.
3. A line segment can be rotated five times that will have it end in its starting position but facing the opposite direction, meaning it rotated 180 degrees, so the angles from a through e are each 180/5 = 36.
My Proof: Draw a pentagon around the star. Each angle of a regular pentagon is 540/5 = 108 degrees. The pentagon I drew makes outer triangles. Each outer triangle is isosceles because it has two sides that are congruent based on the star being regular. Each outer triangle has a 108 degree angle because it is vertical with a 108 degree angle of the pentagon. Therefore the base angles are (180108)/2 = 36 degree each. The outer pentagon is regular, so each corner is 108 degrees. Is it necessary to prove that a pentagon drawn around a regular star will be regular, or can that be assumed? The outer triangles have two sides that are congruent as parts of a regular star, and 108 degree angles as I already proved with vertical angles, so the outer triangles are congruent SAS. Therefore each angle of the outer pentagon has three parts that are congruent by corresponding parts or by angles a through e being congruent. Each corner has an angle in the star I'm solving for and two base angles that I proved were 36 degrees. Therefore each angle I'm solving for is 108  36  36 = 36. I admit that it has a much longer description, but it's what I thought of.
1. The angles from a through e are 36 degrees each because they are inscribed in 72 degree arcs. I don't like circles, so I would never work with them unless I had to.
2. The pentagon has five 108 degree angles, and each triangle has two angles supplementary to 108 degrees, meaning they are 72 degrees each, and angles a through e are 36 degrees each as part of 367272 triangles.
3. A line segment can be rotated five times that will have it end in its starting position but facing the opposite direction, meaning it rotated 180 degrees, so the angles from a through e are each 180/5 = 36.
My Proof: Draw a pentagon around the star. Each angle of a regular pentagon is 540/5 = 108 degrees. The pentagon I drew makes outer triangles. Each outer triangle is isosceles because it has two sides that are congruent based on the star being regular. Each outer triangle has a 108 degree angle because it is vertical with a 108 degree angle of the pentagon. Therefore the base angles are (180108)/2 = 36 degree each. The outer pentagon is regular, so each corner is 108 degrees. Is it necessary to prove that a pentagon drawn around a regular star will be regular, or can that be assumed? The outer triangles have two sides that are congruent as parts of a regular star, and 108 degree angles as I already proved with vertical angles, so the outer triangles are congruent SAS. Therefore each angle of the outer pentagon has three parts that are congruent by corresponding parts or by angles a through e being congruent. Each corner has an angle in the star I'm solving for and two base angles that I proved were 36 degrees. Therefore each angle I'm solving for is 108  36  36 = 36. I admit that it has a much longer description, but it's what I thought of.
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