Square of Sum

Theorem

 * $$\forall x, y \in \R: \left({x + y}\right)^2 = x^2 + 2 x y + y^2$$

Algebraic Proof
Follows directly from the Binomial Theorem:
 * $$\forall n \in \Z_+: \left({x+y}\right)^n = \sum_{k=0}^n {n\choose k}x^{n-k}y^k$$

putting $$n = 2$$.

Geometric Proof
As Euclid put it:


 * "If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments."


 * Euclid-II-4.png

Let the straight line $$AB$$ be cut at random at $$C$$.

Construct the square $ADEB$ on $$AB$$ and join $$DB$$.

Construct $CF$ parallel to $$AD$$ through $$C$$.

Construct $HK$ parallel to $$AB$$ through $$G$$.

From Parallel Implies Equal Alternate Interior Angles, $$\angle CGB = \angle ADB$$.

We have that $$BA = AD$$.

So from Isosceles Triangles have Two Equal Angles, $$\angle ADB = \angle ABD$$.

So $$\angle CGB = \angle CBG$$, and so from Triangle with Two Equal Angles is Isosceles, $$BC = CG$$.

From Opposite Sides and Angles of Parallelogram are Equal, we have $$CB = GK$$ and $$CG = KB$$ and so $$CGKB$$ is equilateral.

Now, since $$CG \| BK$$, from Parallel Implies Supplementary Interior Angles we have that $$\angle KBC$$ and $$\angle GCB$$ are supplementary.

But as $$\angle KBC$$ is a right angle, $$\angle BCG$$ is also a right angle.

So from Opposite Sides and Angles of Parallelogram are Equal, $$\angle CGK$$ and $$\angle GKB$$ are also right angles.

So $$CGKB$$, described on $$CB$$. is right-angled, and as it is equilateral, by definition it is square.

For the same reason, $$HGDF$$ is also square, and described on $$HG$$, which equals $$AB$$.

So $$HGDF$$ and $$CGKB$$ are the squares on $$AC$$ and $$CB$$.

Now we have from Complements of Parallelograms are Equal that $$\Box ACGH = \Box GKEF$$.

But $$\Box ACGH$$ is the rectangle contained by $$AC$$ and $$CB$$, as $$CB = GC$$.

So $$\Box GKEF$$ is also equal to the rectangle contained by $$AC$$ and $$CB$$.

But the squares $$HGFD$$ and $$CBKG$$ are equal to the squares on $$AC$$ and $$CB$$.

So the four areas $$HGFD$$, $$CBKG$$, $$ACGH$$ and $$GKEF$$ are equal to the squares on $$AC$$ and $$CB$$, and twice the rectangle contained by $$AC$$ and $$CB$$.

But $$HGFD$$, $$CBKG$$, $$ACGH$$ and $$GKEF$$ are also equal to the square on $$AB$$.

Hence the result.