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.
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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Converse of Pythagoras Theorem 

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. 
Given : A triangle ABC such that AB square plus BC square equals to AC square. 
To prove : the triangle ABC is right - angled at B. 
Construction: Construct a right - angled triangle PQR, right - angled at Q such that PQ equal to AB and QR equals to BC. 
Prof: Appling the Pythagoras theorem to right - angled triangle PQR we have PQ square plus QR square equals to PR square or AB square plus BC square is equal to PR square since PQ is equal to AB and QR equals to BC by construction. Let this be equation 1 but it is given that AB square plus BC square equals to AC square. 
From equations 1 and 2 we get PR square equals to AC square i.e., PR equals to AC. 
Therefore, by SSS congruence criterion, we get triangle ABC is congruent to triangle PQR so we get angle B equal to angle Q but we know angle Q is equal to 90 degrees by construction. 
Therefore, angle B equals to 90 degrees. 
Hence, triangle ABC is a right triangle, right - angled at B. 
Thus, the theorem is proved.