# Proofs: Hypotenuse is the longest side of a right triangle

• B
Gold Member
I want to prove that the hypotenuse is the longest side of a right angled triangle. Could people check that the proof I'm giving is correct?

Say the hypotenuse is of length ##c## and the other two sides are of length ##a## and ##b##. First of all, we obviously have:

##a^2 + b^2 > a^2 \quad## and ##\quad a^2 + b^2 > b^2##.

By Pythagoras' theorem ##c^2 = a^2 + b^2##. Using this in the above relations we obtain:

Now ##c^2## is the area of a square with sides of length ##c##, and ##a^2## is the area of a square with sides of length ##a##. The relation ##c^2 > a^2## says the square of area ##c^2## is bigger than the square of area ##a^2##, therefore, we must have

##c > a.##

Similarly, ##c^2 > b^2## implies

##c > b##.

And we have proved that the hypotenuse is the longest side.

Q.E.D.

Is this correct? Thanks.

Also, I want to ask what other proofs are there that the hypotenuse is the longest side of a right angled triangle?

Cozma Alex

fresh_42
Mentor
It is correct.
You could use Thales' theorem.

Gold Member
It is correct.
You could use Thales' theorem.
Thanks fresh_42. So I understand Thales' theorem now:

https://en.wikipedia.org/wiki/Thales'_theorem

Given Thales' theorem, in order to complete the proof (that the hypotenuse is the longest side of a right angled triangle) I just need to prove that the diameter is the longest chord of a circle, which I think I know how to do:

Let ##O## be the origin of the circle. Take any chord, say with end points ##A## and ##B##. This chord is denoted ##\overline{AB}##. Then segments ##\overline{AO}## and ##\overline{BO}## are radii of the circle. We have the triangle ##\triangle AOB## (see figure "chord.jpg"). By the triangle inequality:

##
|\overline{AB}| \leq |\overline{AO}| + |\overline{BO}| .
##

Now the diameter, ##D##, is twice the radius which is equal to ##|\overline{AO}| + |\overline{BO}|##. Therefore we have

##
|\overline{AB}| \leq D ,
##

where equality holds only when the triangle ##\triangle AOB## is degenerate which is the case when the chord passes through the origin ##O##.

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fresh_42
Mentor
My thought has been simplier.
Each triangle with a right angle can be chosen to be in a circle, where the hypotenuse is the diameter. (Thales)
And a diameter is the longest secant of all in a circle. Furthermore it has to include the circle's center for that reason. The two other sides of the triangle are secants, too. Since they don't include the center, they can't be diameters and therefore have to be smaller than this.

julian
Svein
I think you are arguing in circles.
• The hypotenuse is defined as the longest side in right angled triangle
• Pythagoras said that "the square of the hypotenuse is equal to the sum of the squares of the cathetes"

Math_QED
Homework Helper
2019 Award
I think you are arguing in circles.
• The hypotenuse is defined as the longest side in right angled triangle
• Pythagoras said that "the square of the hypotenuse is equal to the sum of the squares of the cathetes"
Depends on the definition. You can define it to be the longest side, or you can find something else.

mfb
Mentor
I guess within this thread, the hypotenuse is defined as the side opposite to the 90 degree angle.

The law of sines would be yet another approach.

Gold Member
Depends on the definition. You can define it to be the longest side, or you can find something else.
I'm using the definition that the hypotenuse is the side opposite the right angle.

Gold Member
I guess within this thread, the hypotenuse is defined as the side opposite to the 90 degree angle.

The law of sines would be yet another approach.
You could use the law of sines but you would have to prove that sin A and sin B are less than 1 for angles between 0 and 90 degrees (as we wish to apply it to right angled triangles), you cant just assume the knowledge that they are less than 1 as that defeats the purpose of the exercise (remember ##\sin A = (side \; opposite \; A) / hypotenuse##). You could use:

##
\sin^2 A + \cos^2 A = 1
##

which becomes

##
\sin^2 A + \sin^2 B = 1
##

and then argue from this that we must have ##\sin A < 1## and ##\sin B <1##.

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Gold Member
Actually, the law of sines reminds me of a geometric proof (Euclid) that "The longer side of a triangle is opposite the larger angle":

We can use this to prove that the hypotenuse is the longest side of a right angled triangle. Say the angles are ##A##, ##B##, and ##90^0##. We must have ##A+B =90^0##, which implies ##A < 90^0## and ##B < 90^0##. As such the hypotenuse is longer than the side opposite ##A## and the hypotenuse is longer than the side opposite ##B##. Hence the hypotenuse is the longest side.

fresh_42
Gold Member
Proposition 19 in my previous post refers to proposition 18 which I give the link to:

Gold Member
I think you are arguing in circles.
• The hypotenuse is defined as the longest side in right angled triangle
• Pythagoras said that "the square of the hypotenuse is equal to the sum of the squares of the cathetes"
No. Pythagoras' theorem states:

"In a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides (cathetes)."

Proofs of Pythagoras' theorem use/take the hypotenuse to be the side opposite the right angle.

Additional: Plus fresh_42's proof involving Thales' theorem doesn't use Pythagoras' theorem. He is identifying the hypotenuse with the side opposite the right angle.

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Svein
Gold Member
Not sure what your point is. Thales' theorem doesn't prove Pythagoras' theorem as far as I know.

Here is a proof of Pythagoras' theorem:

https://en.wikipedia.org/wiki/Pythagorean_theorem#/media/File:Pythagoras-proof-anim.svg

It proves that ##c^2 = a^2 + b^2## where ##c## is the length of the side of the triangle opposite the right angle, and ##a## and ##b## are the lengths of the other two sides. Therefore, we have:

"In a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides (cathetes)."

Svein
Not sure what your point is. Thales' theorem doesn't prove Pythagoras' theorem as far as I know.

Here is a proof of Pythagoras' theorem:

https://en.wikipedia.org/wiki/Pythagorean_theorem#/media/File:Pythagoras-proof-anim.svg

It proves that ##c^2 = a^2 + b^2## where ##c## is the length of the side of the triangle opposite the right angle, and ##a## and ##b## are the lengths of the other two sides. Therefore, we have:

"In a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides (cathetes)."
I just wanted to illustrate the fact that the geometric place for the right angle in a right angle triangle is a half circle with the hypotenuse as diameter.

If you want a short proof of Pythagoras' theorem, check out this proof:

Or this one:

Gold Member
I just wanted to illustrate the fact that the geometric place for the right angle in a right angle triangle is a half circle with the hypotenuse as diameter.
Well it is a geometric place for a right angled triangle and the hypotenuse, not sure it is the only geometric setting. Thales' theorem does neatly prove that the side opposite the right angle is the longest side of the right angled triangle - which is what I wanted.

For another geometric setting see the attached file: trigfns.gif - here we take the radius to be the hypotenuse. This would also provide a geometric proof that the hypotenuse is the longest side, as the projection of the hypotenuse onto the ##x-##axis and the ##y-##axis will be less than the radius of the circle (=length of hypotenuse).

Yep, I know these proofs of Pythagoras - my point is in these proofs they take the hypotenuse to be the side opposite the right angle.

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Gold Member
Here is a easy proof. Figure 1 (a) is the right angled triangle where the hypotenuse (side opposite the right angle) is of length ##c## and the other two sides are of lengths ##a## and ##b##. Figure 1 (b) is just the mirror image of this right angled triangle. In figure 2 we join these triangles together to form the isosceles triangle ##\triangle ABC##. Applying the triangle inequality to this isosceles triangle gives:

##
|\overline{AB}| + |\overline{BC}| > |\overline{AC}|
##

or

##
c+c > 2a .
##

In figure 3 we join the right angled triangles together to form the isosceles triangle ##\triangle A'B'C'##. Applying the triangle inequality to this isosceles triangle gives:

##
|\overline{A'B'}| + |\overline{B'C'}| > |\overline{A'C'}|
##

or

##
c+c > 2b .
##

Ta-dah

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Svein