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.
 A: 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 $\;S.S.S.\;$
A: You can use cosine formula as
$BC^2$ = $AB^2 + AC^2 -2.(AB).(AC)cos(A)$
substituting the condition we get $2.(AB).(AC)cos(\angle BAC)=0$
which gives us $\angle$ $BAC =\pi/2$
A: a^2+b^2= c^2 in triangle ABC, and argue towards contradiction as 
Draw the altitude of the triangle towards one of the sides making angle C, and call the length from vertix C to this intersection x. Then 
x^2+h^2= b^2 and (a-x)^2+h^2= c^2. Playing around with these three equations shows you that x= 0 which is a contradiction
A: 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.] 
A: Understand Converse of Pythagoras Theorem with a VIDEO explanation. Click on the link to WATCH the VIDEO: WATCH VIDEO 
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.
