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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 $\Theta=90^\circ$

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    $\begingroup$ It depends what you consider geometric. Is using the Law of Cosines acceptable? $\endgroup$
    – David P
    Sep 2, 2016 at 6:40
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    $\begingroup$ In trig there is the law of cosines which extends this: $a^2 + b^2 - 2ab\cos \theta = c^2$ $\endgroup$
    – fleablood
    Sep 2, 2016 at 6:43
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    $\begingroup$ 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. $\endgroup$
    – Galc127
    Sep 2, 2016 at 7:02
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    $\begingroup$ @fleablood I think your comments answer this question, and would therefore be better posted as an answer. $\endgroup$
    – MvG
    Sep 2, 2016 at 7:18
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    $\begingroup$ 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. $\endgroup$
    – MvG
    Sep 2, 2016 at 7:18

5 Answers 5

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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

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  • $\begingroup$ Pretty, +1. But isn't this geometric (you appeal to SSS)? $\endgroup$
    – rogerl
    Sep 2, 2016 at 19:32
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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.

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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?

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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 $b$.

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Here is a relatively easy to follow algebraic proof: Since we are given c^2 =a^2+b^2 it follows that c is the largest side and since the largest side of a triangle is opposite the largest angle we can thus drop the altitude say h from C to side c which divides side c into segments say m and n so that c=m+n . Now if we assume on the contrary that C is not a right angle then the altitude , say w , from A to the line containing side a is smaller than b i.e. w<b, because the shortest distance from a point to a line is the length of a perpendicular from that point to the line. Now the area of triangle ABC is both (1/2)hc and (1/2)aw so that hc=aw but since w < b this implies ab>hc and thus (a^2)(b^2)>(h^2)(c^2) (I)

Now by the Pythagorean Theorem ( make a diagram with segment n adjacent to side b ) h^2+n^2=b^2 and h^2+m^2=a^2(II) so that using c^2=(m+n)^2=m^2+2mn+n^2 with (II), (I) becomes (h^2+m^2)(h^2+n^2)>(h^2)(m^2+2mn+n^2) which expanded and simplified becomes (h^2-mn)^2>0 so that h^2<>mn (III) We first consider the possibility h^2>mn ( as the possibility that h^2<mn is entirely similar to what follows as we need just reverse all subsequent inequalities) If we use (III) in (II) we get since we are assuming h^2>mn that mn+n^2<b^2 and mn+n^2<a^2 (IIII) If we add the two inequalities in (IIII) we get (m+n)^2<a^2+b^2 but since c=m+n we have the contradiction c^2<c^2 and hence our assumption is false and the Converse of the Pythagorean Theorem is proved.

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