Hi guys! Today I discovered my own proof for the Angle sum property of a triangle. (Or so I think)

The proof goes like this.

1) Take a triangle ABC. Draw an angle bisector AD to the side BC.

2) Let Angle BAD=Angle DAC=\(a^{\circ}\) , Angle ABC=\(b^{\circ}\) and Angle ACB = \(c^{\circ}\)

3) Angle ADC= \(a^ {\circ}+b^{\circ}\) [External angle=Sum of both interior angles]

4) Similarly, Angle ADB = \(a^ {\circ}+c^{\circ}\)

5) \(2a^{\circ}+b^{\circ}+c^{\circ}={180}^{\circ}\) [Angles on a straight line]

6) Angle A+Angle B + Angle C =\({180}^{\circ}\)

Thus proved.

Please leave your comments in the box below. Kindly tell me if this proof has already been derived by someone :)

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## Comments

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TopNewestI would argue that you can't actually "prove" this because the angle sum property is basically the same as the parallel postulate which is independent of the first four postulates. You can even have geometries where the angle sum property does not hold.

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So can we say, That this is valid for the "Commonly used" Geometry? :P

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It is the characteristic axiom of Euclidean geometry. Other geometries that do not have this are called non-Euclidean geometries.

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U used Exterior Angle Property which is proved by ASP. Hence this is wrong. :)

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I assumed we know Exterior angle property :P

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Assuming something in a proof is wrong.

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In step 3, isn't the fact that the external angle is the sum of the other two interior angles actually equivalent to the angle sum property for a triangle? You're saying \(180^{\circ}-\angle ACD=a^{\circ}+b^{\circ}\) since they are the internal angles of a triangle, after all.

Can you alter your proof so that none of the steps assume that the sum of the internal angles of a triangle is \(180^{\circ}\)?

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Well, I had derived this as a solution to a textbook problem when I was of your age. By the way, good work.

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Thank you! :D

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