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.

Well, what always worked for me as a tenth grader was to cut a triangle into right triangles:
Construct a point $D$ on $BC$ so that $DA$ is perpendicular to $BC$ so $ADC$ is a right triangle and $AD^2 + DC^2 = AC^2$.
THere are three possibilities.  
i) $\angle ABC$ is acute.
Then $D$ is between $B$ and $C$ and $BC = BD + DC$.
Then $AB^2 = BD^2 + AD^2; AC^2 = AD^2 + DC^2; BC^2 = BD^2 + 2BD*DC + DC^2$
And $AB^2 + BC^2 = AD^2 + DC^2 + 2(BD^2+ BD*DC) = AC^2 + 2(BD^2+ BD*DC) > AC^2$
(It helps if you draw a picture and note $AB > AD; BC > BD$ so of course $AB^2 + BC^2 > AD^2 + DC^2 = AC$.  But the calculation tells you exactly how much bigger.)
ii) $\angle ABC$ is obtuse.
Then $B$ is between $D$ and $C$ and $DC = DB + BC$.
Then $AB^2 = BD^2 + AD^2; AC^2 = AD^2 + DC^2; DC^2 = DB^2 + 2DB*BC + BC^2$
And $AB^2 + BC^2 = AD^2 + DC^2 - 2(BD^2+ BD*DC) = AC^2 - 2(BD^2+ BD*DC) < AC^2$
iii) $\angle ABC$ is right.
Then $B=D$ and $C$ and $DC =  BC $.
Then $AB^2 =  AD^2; AC^2 = AD^2 + DC^2=AB^2 + BC^2; BC^2 = DC^2 $
So i) $AB^2 + BC^2 > AC^2 \implies \triangle ABC$ is acute and $AB^2 + BC^2 = AC + 2(BD^2 + BD*DC)$.
ii) $AB^2 + BC^2 < AC^2 \implies \triangle ABC$ is abtuse and $AB^2 + BC^2 = AC - 2(BD^2 + BD*DC)$.
iii) $AB^2 + BC^2 = AC^2 \implies  \triangle ABC$ is right.
.... which gives the tenth grader a hint of the law of cosines to come.
[i.e. $\pm(BD^2 + DB*DC) = AB*BC*\cos \angle ABC$ which is $0$ if $\angle ABC$ is right; positive if obtuse and negative if acute.] 
