Area of a Parallelogram: Using congruence and similarity

Given that angle ACB is equal to angle CBE, it follows that AC is parallel to BE

since alternate angles between parallel lines are equal.

As we know that ABCD is a parallelogram, AB is parallel to DC and therefore AB is also parallel to CE since it is a line that continues DC.

Given that AC is parallel to BE and, in addition, AB is parallel to CE, it can be argued that ABCE is a parallelogram and, therefore, each pair of opposite sides in a parallelogram are parallel and equal.

From this it is concluded that AB=CE=9

Now we calculate the area of the parallelogram ABCD according to the data.

$S_{ABCD}=AB\times BF$

We replace the data accordingly:

$S_{ABCD}=9\times6=54$

First, we must remember the formula for the area of a parallelogram:$\text{Lado }x\text{ Altura}$.

In this case, we will try to find the height CH and the side BC.

We start from the side

First, let's observe the small triangle EBG,

As it is a right triangle, we can use the Pythagorean theorem (

$A^2+B^2=C^2$)

$BG^2+4^2=5^2$

$BG^2+16=25$

$BG^2=9$

$BG=3$

Now, let's start looking for GC.

First, remember that the deltoid has two pairs of equal adjacent sides, therefore:$FC=EC=9$

Now we can also do Pythagoras in the triangle GCE.

$GC^2+4^2=9^2$

$GC^2+16=81$

$GC^2=65$

$GC=\sqrt{65}$

Now we can calculate the side BC:

$BC=BG+GT=3+\sqrt{65}\approx11$

Now, let's observe the triangle BGE and DHC

Angle BGE = 90°
Angle CHD = 90°
Angle CDH=EBG because these are opposite parallel angles.

Therefore, there is a ratio of similarity between the two triangles, so:

$\frac{HD}{BG}=\frac{HC}{GE}$

$\frac{HD}{BG}=\frac{7.5}{3}=2.5$

$\frac{HC}{EG}=\frac{HC}{4}=2.5$

$HC=10$

Now that there is a height and a side, all that remains is to calculate.

$10\times11\approx110$