# The Converse of the Pythagoras Theorem

Let $ABC$ be a triangle in the plane. Suppose that $AB^2+AC^2=BC^2.$ Prove that:

$\angle BAC$ is right angle.

Remark: I believe this to be true. Now I have the following difficulty. I'm looking for a proof which a 10th grade student can understand, but any effort of mine is via proof by contradiction. I mean assuming that the given angle is not a right angle and then arrive on a contradiction. So I'm looking for a solution which could avoid the indirect method but if not then a solution by the indirect method.

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We habe $AB^2 + AC^2 = BC^2 + 2AB\cdot BC \cdot \cos(\angle ABC)$ –  martini Aug 16 '13 at 10:59
What about Euclid's proof? It's Proposition 48 in first book, you can find the whole book for free here: farside.ph.utexas.edu/euclid.html –  pppqqq Aug 16 '13 at 10:59
This can be quickly and easily be proved using the law of cosines, yet a 10th grader perhaps hasn't studied that much trigonometry... –  DonAntonio Aug 16 '13 at 10:59
You can use any proof you like of the cosine formula –  Mark Bennet Aug 16 '13 at 11:01
europa.sim.ucm.es/compludoc/… –  Jp McCarthy Aug 16 '13 at 11:13

Draw side by side the given triangle $\;\Delta ABC\;$ and a new triange $\,PQR\;$ as follows:

$$PQ=AB\;,\;\;PR=AC\;,\;\;\angle QPR=90^\circ$$

Now, we're given $\,AB^2+AC^2=BC^2\;$, and Pythagoras theorem also gives $\,PQ^2+PR^2=QR^2\;$ , so

$$BC^2=AB^2+AC^2=PQ^2+PR^2=QR^2\implies BC=QR$$

and we get $\,\Delta ABC\cong\Delta PQR\;$ by $\;S.S.S.\;$

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$BC^2$ = $AB^2 + AC^2 -2.(AB).(AC)cos(A)$
substituting the condition we get $2.(AB).(AC)cos(\angle BAC)=0$
which gives us $\angle$ $BAC =\pi/2$