Trigonometry: Proof concerning congruent triangles

[moriheru]

Homework Statement

Prove that the ratios of the sides of a right angle triangle ( for example hypotenuse divided by ankathete...) are equivalent to the ratios of the congruent triangles.

I believe this problem amounts to showing that sin(alpha)=sin(alpha') and the same for cosinus and tangens, where aplha' is a angle of the congruent triangle

Homework Equations

congruence "laws" (SSS,SAS,SSA,ASA)

The Attempt at a Solution

A triangle is congruent to another if the congruence laws apply ( SSS, SAS, SSA, ASA), so I shall try and prove the statement for each case:
SSS: If all sides are equivalent or simply greater by a specific factor the ratios are the same. So sin(α)=sin(α') if α' is the angle of the congruent triangle. This is also true for cos(α) and tan(α).

SAS: If the angle that is enclosed by the sides is not the right angle then we know all angles seeing as 180-90-α=β
(alpha is the angle we know) and so all angles of both triangles are the same. Sinus, cosinus and tangens only depend on the angle, therefore if all angles are the same sin(α)=sin(α') and the same for cosinus and tangens.

ASA: Again we can deduce all angles and again they must all be the same, so sin(α)=sin(α') and the same for cosinus and tangens

SSA: These are the more tricky ones (assuming everything I have done so far is right).

Let us assume that the side oposite to the angle α (alpha is not the right angle) and the side at the angle aplha are equivalent (where none of the sides are the hypotenuse) and that one of the angles is equivalent for both triangles. Then once again we can deduce all angles and our work is done. In other words we know the tan(α)=tan(α'), sin(α)=sin(α'), cos(α)=cos(α')

If one of the sides is a hypotenuse then it is a bit more complicated. Say the hypotenuse and the side oposite to a angle α (the sinus) are the same for both triangle, then it is clear that sin(α)=sin(α') but not that cos(α)=cos(α') or tan(α)=tan(α'), but one may attempt to use algrebra to show that this is also true: (OS=opposite side)(HYP' is the hypotenuse of the congruent triangle the same for OS')(AS= Side touching the angle)

sin(α)=HYP/OS=HYP'/OS'→ OS'=OS; HYP'=HYP

tan(α)=OS/AS=OS'=AS →AS=AS'→tan(α)=tan(α')

cos(α)=AS/HYP=AS'/HYP'→ cos(α)=cos(α')

So in both cases the above statements are proven true (at least I believe).

I have to hand a proof in by next week so I would be very gratefull if you could read through this rubbish and correct it where it is wrong.

Thank you

 

[mfb]

That is one possible definition of congruent. You cannot prove a definition.

How did you define "congruent triangle"? With those four rules? That would be a problematic and unconventional way.

 

[moriheru]

The proof was if the ratios of the sides of a traingle are the same as the ratios of the sides of a congruent triangle. Wikipedia congruence "Sufficient evidence for congruence between two triangles in Euclidean space can be shown through the following comparisons:" the named rules follow. SO please explain what you meant.

 

[mfb]

The usual logic:

Different shapes are congruent if all corresponding angles and lengths are the same
Using that definition, it is obvious that congruent triangles have the same ratios.

For the special case of triangles, we want to find easier conditions that are sufficient to make them congruent:
Consider two triangles where all three sides are the same. We'll prove that they are congruent: [...]

If you have shown that two triangles are congruent (and therefore that all corresponding sides have the same length), it is trivial that the ratios of sides is the same for both.I am quite sure what you posted as "the full problem statement" is not the actual problem statement, because that would not make sense. It's like "prove that every triangle where the side lengths are 3, 5 and 7 has a side with a length of 5".

 

[moriheru]

I believe you are right. I must restate the problem and then I think it makes sense. My error occurred because I translated from german to english.
The problem should not be that of congruence but of similarity. On https://en.wikipedia.org/wiki/Similarity_(geometry) there are the same or slitly different statements (rules).
If you could tell me if in general my proof, now that I have corrected the statement, is consistent I would be very thankfull.

 

[moriheru]

To restate my ideas and the corrected version of the problem:
There is no problem proving that if all angles and sides are the same the ratios are also equivalent, so I shall just skip them.
I only care about the SSA and SAS cases.

 

[Mark44]

A triangle is congruent to another if the congruence laws apply ( SSS, SAS, SSA, ASA),

That's not the definition of congruence. Two triangles are congruent if (if and only if) the corresponding sides of the triangles are equal and the corresponding angles are equal. As it turns out, it is sufficient for the sides of one triangle to be equal to (congruent to) the corresponding sides of another triangle.

To restate my ideas and the corrected version of the problem:
There is no problem proving that if all angles and sides are the same the ratios are also equivalent, so I shall just skip them.
I only care about the SSA and SAS cases.

The SSA case is a dead end. It's possible for two triangles to have two sides that are congruent, as well as a non-included angle, but the triangles aren't congruent.

Here's a crude sketch of what I'm talking about.

The marked angles above are congruent, the two sides at the left are congruent, and the two sides at the right are congruent. Obviously, the two triangles are NOT congruent.

 

[mfb]

Where is the original translated* problem statement?

*while this is an English forum, if you include the German version (in addition to the English one) it doesn't harm. I am German as well.