Method 1: By similarity:
Prerequisites:
Similarity of triangles (definition)
AA similarity (proof)
Proof:
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Let $\triangle ABC$ be a right triangle, right angled at B. Let us draw an altitude from vertex B to the line AC, meeting AC at a point D.
Now, in $\triangle ADB$ and $\triangle ABC$,
$\qquad\quad\:\!\angle A = \angle A\qquad\qquad\qquad\qquad\quad\:\text{(common)}\\
\qquad\quad\angle ADB = \angle ABC = 90^o\\
\therefore\quad\;\;\;\triangle ADB\sim\triangle ABC\qquad\qquad\qquad\text{(by AA similarity rule)}$
Since, by definition of similar triangles, corresponding sides of similar triangles are proportional,
$\therefore\quad\;\;\dfrac{AD}{AB} = \dfrac{AB}{AC}\\[12pt]
\Rightarrow\quad\;\; AB^2 = AD \times AC\qquad\qquad\qquad\cdots\text{(1)}$
Similarly, from $\triangle BDC$ and $\triangle ABC$,
$\qquad\quad BC^2 = DC \times AC\qquad\qquad\qquad\cdots\text{(2)}$
Adding $(1)$ and $(2)$,
$\qquad\quad\begin{align} AB^2 + BC^2 &= AD\times AC + DC\times AC\\
&= (AD + DC)\times AC\\
&= AC\times AC\\
&= AC^2\end{align}$
Q.E.D.
Method 2: By Area:
Prerequisites:
Square (definition)
Area of square (proof)
Area of right triangle (proof)
SAS congruence (proof)
Angle sum property of triangle (proof)
Angle on a straight line (proof)
$(a+b)^2 = a^2 + b^2$ (proof)
Proof:
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Let ABCD be a square having side length equal to $(a+b)$. Let us locate the points E, F, G and H on the lines AB, BC, CD and DA respectively, such that AE $=$ BF $=$ CG $=$ DH $=$ $a$. Join EF, FG, GH and HE by straight lines, as shown in the figure.
Let $\angle AEH = \theta$, then by angle sum property of a triangle,
$\qquad\quad\begin{align}\angle AHE &= 180^o - \angle A - \angle AEH\\
&= 180^o - 90^o - \theta\\
&= 90 -\theta\qquad\qquad\qquad\qquad\cdots\text{(1)}\end{align}$
Now, in $\triangle HAE$ and $\triangle EBF$,
$\qquad\quad AE = BF = a\qquad\qquad\qquad\text{(by construction)}\\
\qquad\quad AH = BE = b\qquad\qquad\qquad\text{(by construction)}\\
\qquad\quad\angle A = \angle B = 90^o\qquad\qquad\quad\:\!\text{(by definition of square)}$
Hence, $\triangle HAE\cong\triangle EBF$ by SAS rule.
Thus, by CPCTC,
$\qquad\quad\angle BEF = \angle AHE = 90^o - \theta\qquad\qquad\qquad\text{(from $(1)$)}\\
\begin{align}\therefore\quad\;\;\;\angle HEF &= 180^o - \angle AEH - \angle BEF\qquad\quad\text{($\because$ AB is a straight line)}\\
&= 180^o - \theta - (90^o - \theta)\\
&= 90^o\qquad\qquad\qquad\qquad\qquad\qquad\;\cdots\text{(2)}\end{align}$
Similarly, all the four triangles, i.e., $\triangle HAE,\;\triangle EBF,\;\triangle FCG$ and $\triangle GDH$ are congruent. Thus, by CPCTC, $\angle HEF, \; \angle EFG,\; \angle FGH$ and $\angle GHE$ are equal and, from $(2)$, are equal to $90^o$. Also, EF $=$ FG $=$ GH $=$ HE $=$ $c$ (say). Hence, EFGH is a square (see definition of a square).
Further, since these four triangles are congruent, they have same area.
Now, let us find the area of the square ABCD:
$\qquad\quad \text{Area of }\; ABCD = \text{Area of } \;EFGH + ar (\triangle HAE) + ar (\triangle EBF)+ ar (\triangle FCG) \\
\\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad + ar (\triangle GDH)\\
\Rightarrow\quad\;\; (a+b)^2 = c^2 + 4\times\frac{1}{2} ab\qquad\qquad\;\text{(by formula for area of square and right triangle)}\\
\Rightarrow\quad\;\; a^2 + b^2 + 2ab = c^2 + 2ab\qquad\qquad\text{($\because\;\;(a+b)^2 = a^2 + b^2 + 2ab$)}\\
\Rightarrow\quad\;\; a^2 + b^2 = c^2\qquad\qquad\qquad\qquad\quad\;\;\cdots\text{(3)}$
Since, $a$ and $b$ are chosen arbitrarily, hence the result of equation $(3)$ holds for all right triangles.
Q.E.D.
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