Symmetry Destroyed

Here is the alternate geometric approach:

One can observe the symmetry of points $B$ about segment $\overline{CE}$ . Including segment $\overline{CF}$ , we see that these two different-colored triangles have the same areas. We also formed parallelogram $AFCD$ (because $m \angle CFE = m \angle CBE$ ), where there are another pair of different-colored triangles also with same areas.

So because of symmetries, the red and yellow areas are the same.

All three triangles, two yellow and one red, have the same vertical height.

So if the bases can be shown to be the same, the areas will be the same as well.

The combined length of the horizontal bases of the two yellow triangles is $\overline{DC}+\overline{EB}=\overline{FE}+\overline{EB}$

The length of the base of the red triangle is $\overline{AE}=\overline{AF}+\overline{FE}$

But because the trapezoid $ABCD$ is symmetrical $\overline{AF}=\overline{EB}$

So the bases, and therefore the areas as well, are the same.

One way of solving this is to assume values, let $a=5;b=8~and~h=3$ .

from the figure,

$\large a+2x=b$ $\color{#D61F06}\large \implies$ $5+2x=8$ $\color{#D61F06}\large \implies$ $\large 2x=8-5=3$ $\color{#D61F06}\large \implies$ $\large x=1.5$

$\large A_{trapezoid}=\dfrac{1}{2}(a+b)(h)=\dfrac{1}{2}(5+8)(3)=19.5$

$\large A_{red}=\dfrac{1}{2}(b-x)(h)=\dfrac{1}{2}(8-1.5)(3)=$ $\color{#3D99F6}\boxed{9.75}$

$\large A_{yellow}=A_{trapezoid}-A_{red}=19.5-9.75=$ $\color{#3D99F6}\boxed{9.75}$

$\color{plum}\large \boxed{Conclusion:The~area~of~the~red~region~is~equal~to~the~area~of~the~yellow~region.}$

Transform the figure into a rectangle and show that the red and yellow areas are each half of it.

1) Construct a new point F above point A and on the extension of line CD, making a right triangle ADF.

2) Triangle ADF is congruent with EBC. Color ADF yellow; remove EBC. The total area is unchanged since ADF and EBC have the same area; we have simply moved EBC to ADF.

3) The resulting rectangle AECF is divided by its diagonal, AC, into half red and half yellow.

The red and yellow areas are the same.

(Sorry, I found no instructions on how to draw diagrams)

Transfer the yellow region on the right over to the one on the left and you get a rectangle with the top half yellow and the bottom half red. Therefore, the yellow and red regions are equal .

We can use variables and properties of isosceles trapezoids to find the areas of the triangles.

If we cut ∆BEC and paste it next to ∆ADC in such a way that line BC and AD overlap, then we can see that it will become a square with diagonal AC.

Hence, areas of both color will be same.

Note: This is only possible for this case (the square thing) because it is isosceles trapezoid.

Sides AD and CB are equal length so the shape is symmetrical. Triangle CEB can be "removed" and rotated to fit along AD forming a rectangle which is bisected along AC. Areas are therefore equal.

Triangle BCE can be transposed and rotated around its horizontal axis to share side BC with side AD of triangle ADC because the two sides are equal in length. Because angle BEC is 90 degrees, a rectangle is formed with (4) - 90-degree corners, which is bisected by line AC into two equal areas.

Note that the areas of triangles AEC and AED are equal since they have the same height and same base. Now let F be the image of point B when reflected over the perpendicular bisector of EC. Since trapezoid ABC is isosceles it can be shown that the opposite sides of quadrilateral ABFD are congruent, so it is a parallelogram with the same area as trapezoid ABCD. Also, the area of triangle AEB is equal to the area of triangle ACF, so area ACD + area CEB = area ACD + area ACF = area ADF. But EF and AD are parallel, so area AED = area AEF since they have the same base and height. Therefore, area AEC = area ACD + area CEB.

It is obviously true when DC=AB (a rectangle). And also obviously true when DC=0 (a triangle). So if it was not true in general then there would have to be a turning point in between those two extremes. But each of the three regions changes area linearly as you move from one to the other, and you cannot make a turning point from linear functions. Hence there isn't a turning point and the areas are always the same.