Sum of angles in quadrilateral

[shahar]

Why does the sum of the angle is 360 degrees?
Is it consider as theorem?

 

[The Highlander]

Why does the sum of the angle is 360 degrees?
Is it consider as theorem?

I would say it's just a simple fact, rather than a theorem, that the internal angles of any quadrilateral add up to 360° (ie: a complete circle; see below).

(But I suppose it may have been listed as a "theorem" in some ancient Greek mathematician's writings. )

​

 

[R.M.]

Draw quadrilateral ABCD (convex) and then draw diagonal AC. The diagonal splits Angle A into two smaller angles (call them x and y) and also splits angle C into two smaller angles (call them a and b).
Angle B + x + a = 180
Angle D + y + b = 180

Since you have not created any angles inside the quadrilateral, the sum of the angles of these two triangles is equal to the sum of the angles of the quadrilateral.

 

[blamocur]

Why does the sum of the angle is 360 degrees?
Is it consider as theorem?

I consider this a corollary to the Polygon Interior Angles Sum Theorem.

 

[Dr.Peterson]

I would say it's just a simple fact, rather than a theorem

A theorem is a fact that has been proved, is it not?

Is it consider[ed] as theorem?

Yes.

Angle Sum Property of a Quadrilateral | Quadrilaterals Class 8 - GeeksforGeeks

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[The Highlander]

Draw quadrilateral ABCD (convex) and then draw diagonal AC. The diagonal splits Angle A into two smaller angles (call them x and y) and also splits angle C into two smaller angles (call them a and b).
Angle B + x + a = 180
Angle D + y + b = 180

Since you have not created any angles inside the quadrilateral, the sum of the angles of these two triangles is equal to the sum of the angles of the quadrilateral.

Your proof (if that's what you intended it to be) is begging the question (and, therefore, not valid!).
It assumes, as fact. that the interior angles of a triangle add up to 180° but you would have to prove that fact first!

I don't see how your submission (though true) adds anything to what has already been said above (that it is a fact "that the internal angles of any quadrilateral add up to 360°")

However, if we do accept the fact that the internal angles of any triangle add up to 180°, then here is an interesting little video that illustrates how to get the sum of the internal angles of any polygon:-

(NB: Best viewed fullscreen with the sound ON.)

​

 

[The Highlander]

I consider this a corollary to the Polygon Interior Angles Sum Theorem.

Your link doesn't appear to be working.
Perhaps check and re-post?
Thanks.

 

[Aion]

Your proof (if that's what you intended it to be) is begging the question (and, therefore, not valid!).
It assumes, as fact. that the interior angles of a triangle add up to 180° but you would have to prove that fact first!

I don't see how your submission (though true) adds anything to what has already been said above (that it is a fact "that the internal angles of any quadrilateral add up to 360°")

I respectfully disagree. The fact that the angles of a triangle add up to 180° is taken as a given in the proof, but that doesn’t make the argument invalid. In mathematics, especially at the elementary level, it's common to build on well-established results. Every proof starts somewhere, and it's reasonable to assume certain foundational facts when they're already widely accepted and previously proven.

 

[blamocur]

Your link doesn't appear to be working.
Perhaps check and re-post?
Thanks.

Thanks for noticing and notifying.
Hopefully this link works. Also, searching for "Polygon Interior Angles Sum Theorem" provides lots references.

 

[khansaheb]

.... that the angles of a triangle add up to 180°

Actually, I remember that as a theorem from Euclid's book.

In that case, the proposition in #1

the sum of the angle is 360 degrees

Becomes a corollary - at least that is how I was taught in middle school in India (wayyy back when...)

 

[Aion]

I remember reading about this method from Dr Peterson's website. Imagine walking around the perimeter of any convex quadrilateral, say $ABCD$, always turning at the vertices to stay along the sides. Start at point $A$, facing along segment $AB$. At point $B$, turn some angle to now face along $BC$. Repeat at points $C$ and $D$. If you do a full loop and return to your original orientation, the total amount you've turned is $360^\circ$. At each corner, the external turn angle is the supplement of the interior angle, so if we call the interior angles $A,B,C,D$, the external turns are
$$180^\circ-A,180^\circ-B,180^\circ-C,180^\circ-D$$Adding those up, we conclude that

$$4\cdot180^\circ-(A+B+C+D)=360^\circ\implies A+B+C+D=360^\circ$$