# the converse of Pythagoras Theorem

Is it possible to prove the converse of the Pythagoras theorem with out a geometric proof?.That is from $a^2+b^2=c^2$ to $\Theta=90^\circ$

• It depends what you consider geometric. Is using the Law of Cosines acceptable? – David Peterson Sep 2 '16 at 6:40
• In trig there is the law of cosines which extends this: $a^2 + b^2 - 2ab\cos \theta = c^2$ – fleablood Sep 2 '16 at 6:43
• You know that $a^2+b^2=c^2$. From the other hand, you can construct a right angle triangle with lengths $a,b,d$ where $d$ is the hypotenuse. By the Pythagorean theorem $a^2+b^2=d^2$, but this implies that $c^2=d^2$, thus $c=d$. Now the triangles are congruent by SSS, thus their angles are equal, i.e the original triangle is right angled. – Galc127 Sep 2 '16 at 7:02
• @fleablood I think your comments answer this question, and would therefore be better posted as an answer. – MvG Sep 2 '16 at 7:18
• The same holds for @Galc127: I think your comment is in fact an answer (and a nice one at that), and should be posted as such. – MvG Sep 2 '16 at 7:18

The user @MvG noted that it should be an answer rather than a comment.

Let $\triangle ABC$ be some arbitrary triangle with sides $a,b,c$ (assume that $AB=a,BC=b,AC=c$) such that $$a^2+b^2=c^2\tag{1}$$

We can construct $\triangle DEF$ which is right angled and has sides $a,b,d$ (assume $DE=a,EF=b,DF=d$) such that $d$ is length of the hypotenuse, i.e $\angle{DEF}=90^{\circ}$. By Pythagorean theorem we have $$a^2+b^2=d^2\tag{2}$$Now we can put $(1)$ in $(2)$ and get $c^2=d^2$. $c,d>0$, hence $c=d$.

We can use SSS to claim that $\triangle ABC\cong \triangle DEF$, so $\angle{ABC}=\angle{DEF}=90^{\circ}$, hence $\triangle ABC$ is right angled.

Q.E.D

• Pretty, +1. But isn't this geometric (you appeal to SSS)? – rogerl Sep 2 '16 at 19:32

I'm not sure what a non-geometric proof should be. Anything analytic would be based on geometry and circular.

With trigonometry we have the law of cosines that for any triangle with sides $a,b,c$ and the angle $C$ opposite side $c$ we have

$a^2 + b^2 - 2ab\cos C = c^2$

of which $C = 90$ is just a special case. With this, if $C > 90$ then $\cos C < 0$ so $c^2 > a^2 + b^2$ and if $C < 90$ then $c^2 < a^2 + b^2$.

To prove the law of cosines, though, you must assume the Pythagorean Theorem and derive from there.

If you drop an altitude, $h$, to $a$ creating lines $a', a"; a' = a \pm a"$ (plus if $C < 90$; minus if $C > 90$ you will have two right triangles where: $a'^2 + h^2 = c^2$ and $a"^2 + h^2 = b^2$ and $h = b*\sin C$. Trigometric manipulation reveals the law of cosines.

But even without knowing the law of cosines, it's easy to is if $C < 90$ then $a' < a$ and $h < b$ so $c^2 = a'^2 + h^2 < a^2 + b^2$.

And if $C > 90$ then $a' = a + a"$ and $b^2 + a^2 = (a"^2 + h^2) +a^2 < (a" + a)^2 + h^2 = a'^2 + h^2 = c^2$.

But that's all very geometric.

Let we assume that $a,b,c$ are positive numbers and $a^2+b^2=c^2$.

If a triangle has two perpendicular sides with lengths $a$ and $b$, the length of the remaining side is $c$ by the Pythagorean theorem. Assume that a non-right triangle $T$ with side lengths $a,b,c$ exists. By the $SSS$ criterion of congruence, it is possible to overlap such triangle with the previous right triangle, contradiction.

Unwrapped version: by $SSS$, there is a unique triangle with side lenghts $a,b,c$, up to isometries. Since there is a right triangle with such side lengths, every triangle with side lenghts $a,b,c$ is a right triangle.

Alternative, creative version: by Heron's formula, the area of a triangle with side lenghts $a,b,c$ is $\frac{ab}{2}$. That implies the orthogonality of the sides with lenghts $a$ and $b$.

Denote the triangle using vectors $a,b,c$. Then, $$\|c\|^2 = \|a-b\|^2 = (a-b)\cdot (a-b) = \|a\|^2 - 2 a \cdot b + \|b\|^2$$

If $\|a\|^2 + \|b\|^2 = \|c\|^2$ then $a \cdot b = 0$ so $a$ and $b$ are orthogonal.

Note: This is just the law of cosines again but flipped so we define orthogonality using the dot product. Maybe that counts as not geometric?