Congruent triangles - S-S-S theorem

[darkyadoo]

Good afternoon,

First axiom (SAS axiom):
If two sides and the included angle of one triangle are congruent to the corresponding parts of another then the triangles are congruent.

Second axiom :
Let be 3 rays [imath]h,k,l[/imath] on the one hand, and on the other [imath]h',k',l'[/imath] emanating from the same point, respectively.
If [imath]\angle (h,l)\cong \angle (h',l')[/imath] and [imath]\angle (k,l)\cong \angle (k',l')[/imath] then [imath]\angle (h,k)\cong \angle (h',k')[/imath]

How we can prove the following theorem (SSS theorem) using both axioms mentioned above?
if three sides of one triangle are congruent to three sides of another then the triangles are congruent.

I have tried different ways without any success...

 

[darkyadoo]

Let be two triangles [imath]\triangle ABC,\triangle A'B'C'[/imath] such that [math]AB=A'B',BC=B'C',CA=C'A'[/math]
We consider the half-plane [imath]\mathcal S[/imath] delimited by the straight line [imath](A'C')[/imath] in which the point [imath]B'[/imath] doesn't lie upon. Thus there exists a point [imath]H\in \mathcal S[/imath] such that [imath]\triangle A'HC'\cong \triangle ABC[/imath] since [imath]AC=A'C'[/imath].

Then We have two isosceles triangles : [imath]\triangle HA'B', \triangle HC'B'[/imath]so we have [imath]\angle A'HB'\cong \angle A'B'H\; \text{and}\; \angle B'HC'\cong\angle HB'C'[/imath] therefore [imath]\angle A'HC'\cong \angle A'B'C'[/imath] since [imath]\angle A'HC'\cong \angle ABC[/imath] we deduce that [imath]\angle A'B'C'\cong \angle ABC[/imath], then we apply the SAS axiom to conclude that [imath]\triangle ABC\cong\triangle A'B'C'[/imath]

I think that is correct (I've got help on youtube )

 

[Dr.Peterson]

Second axiom :
Let be 3 rays [imath]h,k,l[/imath] on the one hand, and on the other [imath]h',k',l'[/imath] emanating from the same point, respectively.
If [imath]\angle (h,l)\cong \angle (h',l')[/imath] and [imath]\angle (k,l)\cong \angle (k',l')[/imath] then [imath]\angle (h,k)\cong \angle (h',k')[/imath]

As stated, this is false! Here the conditions are true but the conclusion is false:

​

Or are the angles defined as signed angles? I am not familiar with your notation.

In any case, you need to be careful about orientation. This is the source of some false proofs!

Let be two triangles [imath]\triangle ABC,\triangle A'B'C'[/imath] such that [math]AB=A'B',BC=B'C',CA=C'A'[/math]
We consider the half-plane [imath]\mathcal S[/imath] delimited by the straight line [imath](A'C')[/imath] in which the point [imath]B'[/imath] doesn't lie upon. Thus there exists a point [imath]H\in \mathcal S[/imath] such that [imath]\triangle A'HC'\cong \triangle ABC[/imath] since [imath]AC=A'C'[/imath].

Then We have two isosceles triangles : [imath]\triangle HA'B', \triangle HC'B'[/imath]so we have [imath]\angle A'HB'\cong \angle A'B'H\; \text{and}\; \angle B'HC'\cong\angle HB'C'[/imath] therefore [imath]\angle A'HC'\cong \angle A'B'C'[/imath] since [imath]\angle A'HC'\cong \angle ABC[/imath] we deduce that [imath]\angle A'B'C'\cong \angle ABC[/imath], then we apply the SAS axiom to conclude that [imath]\triangle ABC\cong\triangle A'B'C'[/imath]

I think that is correct (I've got help on youtube )

What theorem did you use to prove the existence of H?

Can you explicitly use your second axiom, showing that the orientation of the angles is correct, so that they add up?

And can you exercise your own proof skills, rather than your searching skills?

 

[darkyadoo]

Thanks for your answer, for my searching skill, I wanted an idea to prove this theorem (isosceles triangles) and at this step as we are close to the axioms, I wouldn't find this idea all alone, furthermore they use two transformations (a symmetry and a translation), something I want to avoid, that's why I preferred to assert the existence of that point H, but I still don't have any idea how to prove that theorem without any transformation and using only axioms or other theorems deduced straight from axioms.
For the angle part, I have translated — "LES PRINCIPES FONDAMENTAUX DE LA GEOMETRIE" from M.D. HILBERT — in English which is not my first language.
Anyway thank you for your answer, I'll come back as soon as I'll get the result!

 

[Dr.Peterson]

For the angle part, I have translated — "LES PRINCIPES FONDAMENTAUX DE LA GEOMETRIE" from M.D. HILBERT — in English which is not my first language.

So, you're implying that the problem comes from Hilbert, and says or implies that just those two axioms should be sufficient? You're right that it may be difficult to find a ready-made proof with those constraints (except maybe in Hilbert).

I've never studied Hilbert's system in depth, so I'm not sure of details. But I checked Hilbert's axioms as given here; I see SAS and the angle notation, but I don't see your "angle-addition" axiom. (I presume your book is the French translation of Grundlagen.)

I found this presentation of Hilbert's axioms, which includes a more elaborate angle addition proposition (not axiom). It also includes a proof of SSS soon after that, which looks like the one you found. It uses a different notation.

 

[darkyadoo]

thank you for your answer!

 

[darkyadoo]

So, you're implying that the problem comes from Hilbert, and says or implies that just those two axioms should be sufficient? You're right that it may be difficult to find a ready-made proof with those constraints (except maybe in Hilbert).

I've never studied Hilbert's system in depth, so I'm not sure of details. But I checked Hilbert's axioms as given here; I see SAS and the angle notation, but I don't see your "angle-addition" axiom. (I presume your book is the French translation of Grundlagen.)

I found this presentation of Hilbert's axioms, which includes a more elaborate angle addition proposition (not axiom). It also includes a proof of SSS soon after that, which looks like the one you found. It uses a different notation.

Your document is very helpful, the existence of H is justified by the Corollary 4.1

Good afternoon,

First axiom (SAS axiom):
If two sides and the included angle of one triangle are congruent to the corresponding parts of another then the triangles are congruent.

Second axiom :
Let be 3 rays [imath]h,k,l[/imath] on the one hand, and on the other [imath]h',k',l'[/imath] emanating from the same point, respectively.
If [imath]\angle (h,l)\cong \angle (h',l')[/imath] and [imath]\angle (k,l)\cong \angle (k',l')[/imath] then [imath]\angle (h,k)\cong \angle (h',k')[/imath]

How we can prove the following theorem (SSS theorem) using both axioms mentioned above?
if three sides of one triangle are congruent to three sides of another then the triangles are congruent.

I have tried different ways without any success...

for the second axiom (which is not), the Author may imply that the ray k is between h and l, thus this proposition holds.

Anyway, Now, I'm working with your document which is very interesting. Thank you for your help !