Ggthjy

($ABCD$ is a square. $|BE|=|EC|, |AF|=3cm, |GC|=4cm$. Determine the length of $|FG|$.)

How can I approach this problem, preferably without trigonometry?

(Except proving otherwise, is there a way to know that a given problem cannot be solved without trigonometry?)

Let $H$ be the midpoint of $AC$ and $angle EIC= 90^{circ}$. We can observe that $$FH+3=HG+4,quad FH+HG=x.$$ So we obtain $HG=frac{x-1}2$. Since two corresponding angles are congruent; $angle FEG =angle EHG=45^{circ}$ and $angle EGF=angle HGE$, we have that $triangle FEG$ and $triangle EHG$ are similar to each other. This gives $$FG:EG=EG:HGimplies EG^2 = FGcdot HG=frac{x(x-1)}2.$$ Now, note that $EI=frac14 AC=frac{x+7}4$ and $IG=IH-GH=frac{x+7}{4}-frac{x-1}2=frac{9-x}{4}$. Since $triangle EIG$ is a right triangle, by Pythagorean theorem, we find that
$$
EG^2=frac{x(x-1)}{2}=EI^2+IG^2=frac{(x+7)^2}{16}+frac{(9-x)^2}{16},
$$ which implies $x=5 $ or $x=-frac{13}3$. Since $x>0$, we get $x=5$.

Ok after some calculus, I figured out to solve this problem. Let's draw the vertical V that goes through E, and call $alpha$ and $frac{pi}{4}-alpha$ the angles we get on the left and right of the 45° angle cut by V and call c the length of the side of the square. We can then do some trigonometry to get :

• 
  $c cdot tan(alpha) - (c-frac{3}{sqrt{2}}) = c cdot tan(alpha)$ (Continue the line EF to cut AD)


• 
  $ tan(frac{pi}{4}-alpha) = c cdot frac{sqrt{2}}{8}-1$ (G is $frac{4}{sqrt{2}}$ away vertically from the right side and $frac{c}{2}-frac{4}{sqrt{2}}$ away horizontally from the middle)

With a little trigo, we have 2 equations in $c$ and $tan(alpha)$ that we can solve, which allow us to solve the whole problem. This makes me think that there is no "simple" answer without trigo :(

Calling

$$
AF = a\
FG=x\
GC = b\
EC = c
$$

Considering in the plane $Xtimes Y$ the square diagonal as the $X$ axis with $A$ at the origin, we have

$$
frac{sqrt 2}{2}(a+x+b)= 2c\
x = sqrt2 r\
left(X_0-left(a+frac x2right)right)^2+left(Y_0-frac x2right)^2= r^2\
X_0 = a+x+b -frac{sqrt 2}{2}c\
Y_0 = frac{sqrt 2}{2}c
$$

Here $r$ represents the radius for the circle which intersects the $X$ axis at $F, G$ such that $angle FCG = frac{pi}{4}$ is the angle subtended by arc $AB$

Solving for $x,r,X_0,Y_0$ we get at

$$
left{
begin{array}{rcl}
x& =& frac{1}{3} left(b-a+2 sqrt{a^2-2 b a+4 b^2}right) \
c& =& frac{a+2 b+sqrt{a^2-2 b a+4 b^2}}{3 sqrt{2}} \
r& =& frac{b-a+2 sqrt{a^2-2 b a+4 b^2}}{3 sqrt{2}} \
X_0&=&frac{1}{2} left(a+2 b+sqrt{a^2-2 b a+4 b^2}right) \
Y_0&=&frac{1}{6} left(a+2 b+sqrt{a^2-2 b a+4 b^2}right) \
end{array}
right.
$$

giving after substitution $x = 5$

Solution without trigonometry:
Point G and F can be used to partition your figure into trapezoids by drawing a horizontal line through them. Then you can add up the areas of the figures formed and setting them equal to the area of the square which is side times side.