"Sum of Interior Angle Theorem of a Triangle" PPT Courseware 2

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"Sum of Interior Angle Theorem of a Triangle" PPT Courseware 2

Practical operation

Is there any way you can verify this?

Can you think of a way to prove it from the process of fighting just now?

Proof 1: The sum of the interior angles of a triangle is equal to 180°.

Extend BC to D, outside △ABC, with CA as one side,

CE is the other side as ∠1=∠A,

Therefore CE∥BA (the internal offset angles are equal and the two straight lines are parallel).

∴∠B=∠2 (the two straight lines are parallel and have equal angles).

∵∠1+∠2+∠ACB=180°

∴∠A+∠B+∠ACB=180°

Proof 2: The sum of the interior angles of a triangle is equal to 180°.

Extend BC to D, cross C and make CE∥BA,

∴ ∠A=∠1 (the two straight lines are parallel and the internal offset angles are equal)

∠B=∠2 (two straight lines are parallel and have equal angles)

∵∠1+∠2+∠ACB=180°

∴∠A+∠B+∠ACB=180°

From the picture on the right and the sum of the interior angles of a triangle theorem, what else did you discover?

From ∠ ACE=∠A, ∠ECD= ∠B

It can be seen that ∠ ACD=∠A+ ∠B;

∠ACD >∠A, ∠ACD >∠B.

Corollary 1 An exterior angle of a triangle is equal to the sum of its two non-adjacent interior angles.

Corollary 2 An exterior angle of a triangle is greater than any interior angle that is not adjacent to it.

Expand and extend

Estimate the degrees of ∠A, ∠B, ∠C, ∠D and ∠E in the regular five-pointed star, guess what their sum is, and prove your conjecture.

In △GCE, from Corollary 1, we get ∠1=∠C+∠E

In the same way, ∠2=∠B+∠D

In △AGH, according to the triangle interior angle sum theorem, ∠A+∠1+∠2=180 degrees

So ∠A+∠B+∠C+∠D+∠E=180 degrees

So the sum of the five angles of the pentagram is 180 degrees.

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Update Time: 2024-10-08

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