Triangulo loco

Let $\angle BEC=\alpha$ :

$\sin(\alpha)=\frac{2}{\sqrt{5}}$

$\angle FBG=90^{\circ}-\angle EBC=90^{\circ}-(90^{\circ}-\alpha)=\alpha$

$EF=FC \Rightarrow \angle EFC=\angle GFB=90^{\circ}-\frac{\alpha}{2}$

$\angle FGB=180^{\circ}-\angle FBG-\angle FGB=90^{\circ}-\frac{\alpha}{2}=\angle GFB \Rightarrow FB=BG$

$EF=EC=1, EB=\sqrt{1^2+2^2}=\sqrt{5} \Rightarrow FB=BG=EB-EF=\sqrt{5}-1$

$\Delta(FBG)=\frac{1}{2} \times \sin(\alpha) \times FB \times BG=\frac{1}{2} \times \frac{2}{\sqrt{5}} \times (\sqrt{5}-1)^2$

$\cdots=\frac{\sqrt{5}}{5} \times (6-2 \sqrt{5})=\frac{6 \sqrt{5} -10}{5}$

$a=6,b=5,c=10,d=5 \Rightarrow a+b+c+d=\boxed{\boxed{26}}$

In rt. triangle CEB, ED= $\sqrt5$ and SinCEB= $\dfrac 2 {\sqrt5}$ .

Area of isosceles triangle CEF= $\frac 1 2 *SinCEB * 1^2=\dfrac 1 {\sqrt5}.$

Since GB | | CE, triangles GBF and CEF are similar and FB:EF::( $\sqrt5$ -1):1.

$So\ area\ \Delta\ FGB=area\ \Delta\ CEF*(\dfrac{FB}{EF})^2=\dfrac 1 {\sqrt5}* (\sqrt5 -1)^2. \\ =\dfrac{6-2*\sqrt5}{\sqrt5}=\dfrac{6\sqrt5 - 10} 5=\dfrac{a\sqrt b - c}d. \\ a+b+c+d=6+5+10+5=26.$

Let the perpendicular from $F$ to $CD$ intersect this side at $P$ . Then $\Delta EFP$ is similar to $\Delta EBC$ , and so

$\dfrac{|FP|}{|EF|} = \dfrac{|BC|}{|EB|} \Longrightarrow |FP| = \dfrac{2}{\sqrt{5}}$ , since $|EF| = 1$ and $|EB|^{2} = |EC|^{2} + |CB|^{2}$ .

The area of $\Delta FCE$ is then $\dfrac{1}{2}|EC||FP| = \dfrac{1}{2} \times 1 \times \dfrac{2}{\sqrt{5}} = \dfrac{1}{\sqrt{5}}$ .

Now note that triangles $\Delta FCE$ and $\Delta FGB$ are similar. Letting $Q$ be the point on $AB$ where the perpendicular from $F$ intersects that side, we have that

$\dfrac{|FQ|}{|FP|} = \dfrac{2 - |FP|}{|FP|} = \dfrac{2 - \frac{2}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = \sqrt{5} - 1$ ,

and thus the area of $\Delta FGB$ will be $(\sqrt{5} - 1)^{2}$ that of $\Delta FCE$ , since by similarity we will have $|GB| = (\sqrt{5} - 1)|CE|$ as well. Thus the desired area will be

$(\sqrt{5} - 1)^{2} \times \dfrac{1}{\sqrt{5}} = \dfrac{6 - 2\sqrt{5}}{\sqrt{5}} = \dfrac{6\sqrt{5} - 10}{5}$ .

This results in a final answer of $a + b + c + d = 6 + 5 + 10 + 5 = \boxed{26}$ .

ΔFGB is similar to ΔEFC. By trigonometry,sin <CEF =2/√5, The area [EFC] = 1/√5, The area ratio is the square of length ratio,
(1/√5)/[FBG] =(〖1/(√5-1))〗^2, So the area of triangle FBG, [FBG] = (6√5-10)/5 , a+b+c+d = 26

Use ratio of areas of triangles sharing the same altitude and the ratio of areas of similar triangles.