Two right triangles share a side as follows:

Triangle ABD is isosceles.....and angle BAD   = 45

So angle DAC = 45....so....angle CAB is bisected....which sets up the following relationship :

CA / AB  = CE / BE

But triangle ABC is a right triangle   ...so   BC   =  sqrt (10^2 + 6^2)   =  sqrt (136)  = 2 sqrt (34)

So

10  / 6  = CE / BE

So....CB  is divided into 16 equal parts....and  CE  is 10 of these....so  CE  = (10 / 16) * 2 sqrt (34)  =

(5/4) sqrt (34)  = 1.25 sqrt (34)

The measure of angle  ACB  =   arcsin (6/(2sqrt (34)) 

So....the area of triangle ACE  = (1/2)(10)(1.25)sqrt (34)sin (arcsin (6/(2sqrt (34)) )  =

(1/4) * 10 * 1.25 * 6   =     

18.75  units^2

Here's an alternative method.

Call the distance of E from the line AC, x, then the distance of E from the line BD will be 6 - x.

Call the distance of E from the line AB, y.

We can now write down the areas of the three smaller triangles.

Area ACE = 5x, area AEB = 3y and area BED = 3(6 - x).

So, area ACB = 30 = 5x + 3y, and area ABD = 18 = 3(6 - x) + 3y.

From the second equation, x = y, so substituting into the first equation 8x = 30, x = 15/4.

So, area ACE = 5*(15/4) = 75/4, (which is the one we are asked to find), area AEB = 3*(15/4) = 45/4

and area BED = 3(6 - 15/4) = 27/4.