How to "coordinate bash" a geometry problem? How do you perform "coordinate bashing"? I read through this which gave some useful information, but I still don't know how to perform it. What I know about coordinate bashing:

  
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*Coordinate bashing is assigning geometrical figures points on the coordinate plane.
  
*It involves using formulas such as the distance formula, slope formula, shoelace theorem, point to line distance, etc.
  

Can someone provide an illustrated example or two?
 A: Basically, in order to prove a geometric theorem, you just put the situation on a Euclidean plane, and do the algebra. For example, here's a proof of the law of cosines: 
Say we have a triangle with vertices at $A=(0,0), B=(0, b),$  and $C=(c_1, c_2)$, where $b,c_1 > 0$. Note that this case is general, since you can always rotate/translate a triangle so that it lies on the right half of the plane with any vertex at the origin, and a side lying on the positive $y$-axis. The angle at $A$ is just $\theta = \pi/2 - \tan^{-1}(c_2/c_1)$. Then $\cos(\theta) = \frac{c_2/c_1}{\sqrt{1+(c_2/c_1)^2}}$. Then we have 
$$\overline{AB}^2+\overline{AC}^2-2\overline{AB}\overline{AC}\cos(\theta) = $$
$$b^2+c_1^2+c_2^2-2b\sqrt{c_1^2+c_2^2}\cdot\frac{c_2}{\sqrt{c_1^2+c_2^2}} = $$
$$(c_2-b)^2+c_1^2 = \overline{BC}^2$$
A: Notation: Let $\triangle (PQR)$  denote the area of triangle PQR. Let $|ST|$ denote the length of the line segment $ST.$ 
Let $A,B,C$ be the vertices of a triangle. Let $C'$ be a point on $AB$ and $B'$ be a point on $AC,$ with $C'$ strictly between $A,B,$ and $B'$ strictly between $A,C.$
Let $0<x<1$. Let $C''$ lie on $C'C$ and let $B''$ lie on $B'B$ such that $$|C'C''|=x|C'C| \quad \text { and }\quad |B'B''|=(1-x)|B'B|.$$  Theorem: The ratio of $\triangle (AB''C'')$ to the area of the quadrilateral $BC'B'C$ is $x(1-x).$ 
Proof: Choose orthogonal co-ordinate axes with $A=(0,0).$ Let $p\times q$ denote the scalar outer product. That is, $(a,b)\times (c,d)=ad-bc.$ We use the following tools:
(1).  The area of any triangle $APQ$ is $\frac {1}{2}|P\times Q|.$ 
(2). The basic rules for the outer product: (i).$ u\times v=-(v\times u) .$ (ii). $u\times u=0.$ (iii). For real $r,s$ and vectors $u,v$ we have $(ru)\times v=u\times rv=r(u\times v),$ $(rs)(u\times v)=r((su)\times v)=(rsu\times v),$ and $(r+s)(u\times v)=(ru)\times v +(su)\times v.$
Now let $B'=y C$ and $C'= z B$ with $y,z \in (0,1).$ Then $\triangle (AB'C')=\frac {1}{2}yz |B\times C| =yz \triangle (ABC).$ The area of quadrilateral $BC'B'C$ is therefore $$\triangle (ABC)-\triangle (AB'C')=(1-yz)\triangle (ABC).$$
We have $B''=xB'+(1-x)B$ and $C''=(1-x)C'+xC.$  Plug the values $B'=yC$ and $C'=zB$ into these and compute $\triangle (AB''C'')=\frac {1}{2}|B''\times C''|.$ The calculation effortlessly simplifies to $$\frac {1}{2}x(1-x)(1-yz)|B\times C|=x(1-x)\cdot (1-yz)\triangle (ABC).$$ 
For a proof without co-ordinates, prove that the line thru $C''$ parallel to $AB,$ and the line thru $B''$ parallel to $AC$ both meet the segment  $BC$ in a common point $D.$ Subdivide $ABC$ into 6 triangles,one of which is $AB''C"$ and one is $B''C''D,$ compute the areas of each of the 5 triangles other than AB''C''  and subtact their total from $\triangle (ABC)$ to find $\triangle (AB''C'').$  A complication is that, reading the vertices of triangle $AB''C''$ counter-clockwise from $A,$ they may be in that order, or in the order $A,C'',B''. $
