A geometry problem by Isaac Arce Aguilar

Here is an image for better understanding of my solution:

First we need to know the length of the segment $CF$ , by Pythagorean Theorem we obtain:

$DF^2+DC^2=CF^2=2^2+1^2=5$ $\rightarrow CF=\sqrt{5}$

Now we need to now at what distance is the point $G$ from points $F$ and $C$ , for that we can call $x$ and $y$ the length $CG$ and $FG$ respectively, then we can establish a system of equations:

$(1)$ $x+y=\sqrt{5}$

$(2)$ $2x=3y$

Clearing $x$ in Eq $2$ and substituting it in Eq. $1$ we get:

$\dfrac{3y}{2}+y=\sqrt{5}$

$3y=2(\sqrt{5}-y)$ $\Rightarrow3y=2\sqrt{5}-2y$

$5y=2\sqrt{5}$ $\Rightarrow y=\dfrac{2\sqrt{5}}{5}$

Solving for $x$ we get: $x=\dfrac{3\sqrt{5}}{5}$

Now, like in the image we can draw a segment parallel to $DF$ calling the intersection with $DC$ , $I$ , now due to the simirality of $\triangle{FDC}$ and $\triangle{IGC}$ using Tales Theorem we can calculate the length of $IC$ :

$\dfrac{2}{y}=\dfrac{\sqrt{5}}{\dfrac{3}{5}\sqrt{5}}$ $\Rightarrow\dfrac{2}{y}=\dfrac{1}{\dfrac{3}{5}}=\dfrac{5}{3}$

$6=5y$ $\Rightarrow IC=y=\dfrac{6}{5}$

So, the length of $IG$ :

$IG^2=(\dfrac{3}{5}\sqrt{5})^2-(\dfrac{6}{5})^2=\dfrac{45}{25}-\dfrac{36}{25}=\dfrac{9}{25}$ $\Rightarrow IG=\dfrac{3}{5}$

Now, we can almost calculate the area of $\triangle{GEB}$ , for that we can extend the segment $IG$ intersecting with $AB$ at $J$ , and a parallel segment to $IC$ intersecting at point $G$ and with $CB$ at $H$ , clearly $GHJB$ it's a square. The area $\triangle{GHB}$ its half of the area of the square and $\triangle{GJB}\cong{\triangle{GBH}}$ . Since $IG=CH$ the length of $HB=2-\dfrac{3}{5}=\dfrac{7}{5}$

So the area of $\triangle{BEG}$ could be calculated substracting the area of $\triangle{GJE}$ from the area of $\triangle{GJB}$ :

$Area$ $of$ $\triangle{GJB}=(\dfrac{6}{5}\times\dfrac{7}{5})\div{2}$ $\Rightarrow \dfrac{42}{25}\div{2}=\dfrac{42}{50}=\dfrac{21}{25}$

Since the length of $EB$ is $1$ and $JB$ equals to $\dfrac{6}{5}$ , the length of $JB$ is: $\dfrac{1}{5}$ . So, its area is:

$(\dfrac{7}{5}\times\dfrac{1}{5})\div{2}=\dfrac{7}{25}\div{2}=\dfrac{7}{50}$

Finally, the area of $\triangle{BEG}$ is:

$\dfrac{21}{25}-\dfrac{7}{50}=\dfrac{42}{50}-\dfrac{7}{50}=\dfrac{35}{50}=\boxed{\dfrac{7}{10}}$

Draw two lines parallel to $AB$ and $CD$ through $G$ and $F$ and they meet $BC$ at $H$ and $I$ respective.

Because $\triangle GCH$ and $\triangle FCI$ are similar,

$\dfrac {CH}{CI} = \dfrac {CG}{CF} = \dfrac {CG}{CG+FG} = \dfrac {\frac {3}{2}FG}{(1+\frac {3}{2}) FG} = \dfrac {3}{5} \quad \Rightarrow CH = \frac {3}{5} CI = 0.6$

Area of $\triangle BEG = \frac {1}{2} \times BE \times BH$ $\quad \quad \quad \quad \quad \quad \space = \frac {1}{2} \times BE \times (BC - CH) = \frac {1}{2} \times 1 \times (2-0.6) = \boxed {0.7}$

use similarity of triangles to find the height of triangle BEG and then apply 1/2 base height and find the area,

1/2 * 7/5*1 = 7/10 = 0.7

The distance from G to DC can be found by Thales' Theorem = CG/CF = 3/5. SO, the distance from G to EB is 2- 3/5 = 7/5. So, the area of BEG is 1/2 * 1* 7/5 = 7/10.

Easy Question btw :D You guys complex it Just use similarity to find XG where X belong to CD then 2 - XG = triangle BEG height = 2-0.6=1.4 area triangle BEG = 1/2 x base x height = 1.4 / 2 = 0.7 cm^2

Here, CF= sqrt(FD^2+CD^2)=>CF= sqrt(1^2+2^2)[FD= 0.5XAD=> FD= 0.5X2=> FD= 1]=>CF= sqrt(5)

Now, CG/FG= 3/2=> (CG+FG)/FG= (3+2)/2=> CF/FG= 5/2=>sqrt(5)/FG= 5/2=> FG= 2/sqrt(5)

Now, CG/FG= 3/2=> CG= (3/2)XFG=> CG= (3/2)X(2/sqrt(5))=> CG= 3/sqrt(5)

Now, imagine, we draw a perpendicular from point G upon lines AB and CD.Let, perpendicular drawn from G touches line AB at point I and perpendicular drawn from G touches line CD at point H.

Now, right triangle CFD and right triangle CGH are similar [<CDF= < CHG= 90 and <FCD= <GCH (common angle). So, <FDH= <GHC]

So, we can write, FD/GH= CF/CG=> GH= FDX(CG/CF)=> GH= 1X((3/sqrt(5))/sqrt(5))=> GH= 3/5

Now, GI= HI- GH= 2- (3/5)= 7/5

Finally, area of triangle BEG= 0.5XbXh = 0.5XBEXGI = 0.5X1X(7/5) [As, BE= 0.5XAB= 0.5X2= 1] = 7/10