# (Proof) Two right triangles are congruent.

• B
• vantroff
In summary, Serge Lang's question is about proving the congruence of two right triangles with corresponding sides of equal length. The first attempt uses the Pythagorean theorem and Theorem 10 from the book, while the second attempt uses isometries. The second attempt is highlighted as more important to the person asking the question. Both attempts are valid and the red part can be replaced with the blue part, or omitted altogether.
vantroff
Hi, the question is from Serge Lang - Basic mathematics, Page 171 exercise 6.
Thing to prove:
Let ΔPQM and ΔP'Q'M' be right triangles whose right angles are at Q and Q', respectively. Assume that the corresponding legs have the same length:
d(P,Q)=d(P',Q')
d(Q,M)=d(Q',M')
Then the right triangles are congruent.

My first attempt by using the just the Pythagorean theorem.

d(Q,M)2 + d(P,Q)2 = d(P,M)2
d(Q,M)2 + d(P,Q)2
= d(Q',M')2 + d(P',Q')2 = d(P',M')2

It follows that:
d(P,M)2 = d(P',M')2
and
d(P,M) = d(P',M')

We conclude that the triangles are congruent by Theorem 10 in the book (page 168) which states the following:

Let ΔP,Q,M and ΔP',Q',M' be triangles whose corresponding sides have the equal length, that is
d(P,M)=d(P',M')
d(P,Q)=d(P',Q')
d(Q,M)=d(Q',M')
These triangles are congruent.

Second attempt by using isometries (not using Theorem 10 from the book).

There exist translation which maps Q on Q'. Hence it suffices to prove our assertion when Q=Q'. We now assume Q=Q'. Since d(Q,M)=d(Q,M') there exist rotation relative to Q which maps M on M'. This rotation leaves Q fixed.
This reduces the case to :
Q=Q'
M=M'

Now either P=P' or P≠P'. We assume P≠P'.
Let LQM be the line passing through point Q and M.
The segments P'Q and PQ are perpendicular to LQM. Also, d(Q,P')=d(Q,P).
Q lies on the perpendicular bisector of the segment PP'. The only perpendicular line to PP' which passes through Q is LQM. We conclude that LQM is the perpendicular bisector of PP' and that poins P,Q,P' are collinear and P is at the same distance from Q as P' but in the opposite direction.

There exist reflection through LQM which leaves LQM fixed and maps P on P'.

We found that there exist composite isometry F such that,

F(Q)=Q'
F(M)=M'

F(P)=P'

and given that the image of line segment PQ under isometry G is the line segment between G(P) and G(Q), we conclude the assertion proven.

Can the red part be replaced by this blue part (or omitted)?
Angle ∠MQP=90° and ∠P'QM=90° ⇒ ∠ P'QP is 180° ⇒ P,Q,P' are collinear.

The second proof is more important to me because I'm at the Isometries part of the book. Is there something wrong with these proofs?
Thanks in advance for everyone who will spend time on this.

Last edited:
Looks fine, both with red and blue. With the blue text, I would keep the last part of the red text about the same distance.

vantroff
Just using the fact that my post is at the top...

## 1. What does it mean for two right triangles to be congruent?

Two right triangles are congruent if they have the same size and shape. This means that all corresponding sides and angles are equal. When two triangles are congruent, they are essentially identical and can be superimposed on top of each other.

## 2. How can you prove that two right triangles are congruent?

There are several ways to prove that two right triangles are congruent. One method is using the Side-Angle-Side (SAS) postulate, which states that if two sides and the included angle of one triangle are congruent to the corresponding sides and included angle of another triangle, then the triangles are congruent.

## 3. Why is it important to prove that two right triangles are congruent?

Proving that two right triangles are congruent is important in geometry because it allows us to make conclusions about their corresponding sides and angles. It also helps us to solve problems involving similar shapes or constructions.

## 4. Can two right triangles with different side lengths be congruent?

No, two right triangles with different side lengths cannot be congruent. For two triangles to be congruent, all corresponding sides and angles must be equal. Therefore, if two triangles have different side lengths, they cannot be congruent.

## 5. Are there any shortcuts to proving that two right triangles are congruent?

Yes, there are several shortcuts or postulates that can be used to prove that two right triangles are congruent. These include the Angle-Angle-Side (AAS) postulate, the Hypotenuse-Leg (HL) theorem, and the Side-Side-Side (SSS) postulate. These shortcuts can be helpful in certain situations, but it is important to understand and use the appropriate one for each specific problem.

• General Math
Replies
9
Views
571
• General Math
Replies
2
Views
2K
• Calculus
Replies
9
Views
983
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Science and Math Textbooks
Replies
10
Views
2K
• General Math
Replies
43
Views
5K
• General Math
Replies
2
Views
2K
• General Math
Replies
2
Views
861
• Math Proof Training and Practice
Replies
10
Views
1K
• Classical Physics
Replies
4
Views
751