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 :)

## 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. – Mursalin Habib · 1 year, 11 months ago

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– Mehul Arora · 1 year, 11 months ago

So can we say, That this is valid for the "Commonly used" Geometry? :PLog in to reply

– Mursalin Habib · 1 year, 11 months ago

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

U used Exterior Angle Property which is proved by ASP. Hence this is wrong. :) – Rajdeep Dhingra · 1 year, 11 months ago

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– Mehul Arora · 1 year, 11 months ago

I assumed we know Exterior angle property :PLog in to reply

– Rajdeep Dhingra · 1 year, 11 months ago

Assuming something in a proof is wrong.Log in to reply

– Mehul Arora · 1 year, 11 months ago

I'm just joking :PLog in to reply

– Rajdeep Dhingra · 1 year, 11 months ago

Hey come on Hangouts.Log in to reply

– Mehul Arora · 1 year, 11 months ago

Okay :)Log in to reply

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 \(180^{\circ}\)? – Maggie Miller · 1 year, 11 months ago

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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. – Satyajit Mohanty · 1 year, 11 months ago

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– Mehul Arora · 1 year, 11 months ago

Thank you! :DLog in to reply