Geometrical Proof of a Rotation I wanna prove geometrically ( and not by linear algebra, doing transformations in the  bases ) the result of the rotation of a point. The proof should only include geometrical steps like using  similarity between triangles, pythagorean theorem and definition of cos and sen on a triangle, for example.
This picture clears my doubt    :
http://postimg.org/image/z60zv5d83/
Thanks in advance.
 A: Let $(x,y)$ be the original coordinates, and $(x',y')$ the later one. Then $$x'=ax+by,~y'=cx+dy$$as they are linear combination. Notice that $$x^2+y^2=(x')^2+(y')^2$$Replacing $x',y'$ by $x,y$ in the above equation, we get $$a^2+c^2=b^2+d^2=1,~ab+cd=0$$Let $a=\cos\theta$ and do some work gives the result.

As required by OP, I will give an alternative method using vector decomposition:

Let $\vec{OA}=(x,y)$ be the original vector, $\vec{OB}=(x',y')$ the vector after rotation of angle $\theta$. Then 
$$(x',y')=\vec{OB}=\vec{OC}+\vec{OD}=(x,y)\cos\theta+(y,-x)\sin\theta.~~\mbox{(DONE!)}$$
A: Oh, i just thought of an easy solution  :     
http://postimg.org/image/o4lvhlg57/
But i just wanted some way to understand whats happening in order to be able to write the formulas steadily witouth memorizing. 
The formula says that the point suffers a dcos(theta ) - fsen(theta) horizontal displacement for example. I wanted to know intuitively why thats true.
A: No transformation of bases is needed. Let $R:=R_\theta$ be the rotation about the origin by angle $\theta$.
All is needed, that this is a linear transformation. This can be viewed as a purely geometrical statement:


*

*$0$ goes to $0$.

*Rotation preserves line segments and takes triangles to triangles (corresponding to $R(a+b)=R(a)+R(b)$ for vectors $a,b$).

*Rotation is exchangable with all zooms from the origin, i.e. $R(\lambda a)=\lambda R(a)$.


Let $e_1=(1,0)$ and $e_2=(0,1)$. Now, by definition of sine and cosine, we have that $R(e_1)=(\cos\theta,\sin\theta)$, and please check (geometrically, i.e. by drawing) that $R(e_2)=(-\sin\theta,\cos\theta)$.
Then, for our arbitrary point $P=(d,f)$, rotating it about $0$ is the same as rotating the $\vec{OP}$ vector, so by linearity, we have
$$R(P)=R(d\cdot e_1+f\cdot e_2)=d\cdot R(e_1)+f\cdot R(e_2)\,,$$
and we're there.
