Math Problem need helppppp

Here's one way to do this

Here's a pic :

DE/AD = CB/AB

DE/4.5 =13/9

DE = 13(4.5)/9  = 6.5

So ....the area of triangle ADE = (1/2)(DE*AD = (1/2)6.5 * 4.5 = 14.625 units^2

And we can find X  as the intersection of two lines

The first has the equation  y = (4.5 /6.5)x   ⇒ y = (9/13)x    (1)

D= (0, 4.5)  C = (13, 0)   so the slope of the line connecting these two points is

-4.5 /13  = -9/26

So....the line through these these two points is y = (-9/26)x + 4.5     (2)

Setting (1) = (2)  to find the x cooordinate of X we have

(9/13)x = (-9/26)x + 4.5

[(18/26) + (9/26) ] x = 4.5

(27/26)x = 9/2

x = (26/27)(9/2)  = 13/3 

And y = (9/13)(13/3) = 3

So.....the area of triangle DXE = (1/2) (4.5 - 3)(6.5) = (1/2) (1.5)(6.5) = 4.875 units^2

So the area of AEXD = area of triangle ADE + area of triangle DXE =  14.625 + 4.875 = 19.5 units^2

And the area of triangle BXC = (1/2)(13)(3) = 39/2 = 19.5 units^2

So  the ratio of the areas =  19.5 / 19.5  =   1    ..... exactly the same areas!!!

Here's another way to do this with similar triangles....it is much easier

First....calculate DE as I did before = 6.5

Note that angle BXC = angle DXE  (vertical angles)

Angle DEX = Angle CBX   (alternate interior angles between  parallels )

So  triangle BXC is isimilar to triangle EXD

Since base ED is 1/2 of BC

Then the height of triangle EXD is 1/2 that of triangle BXC

So y = 4.5 is split into  tree equal parts....and the height of BXC is (2/3)of these = 3 

Then  area of triangle EXD = 1/4 area of triangle BXC

Area of BXC = (1/2)(3)(13) =  39/2  = 19.5

And the area of EXD = (39/8)

And the area of triangle ADE is (1/2)(4.5)(6.5) = (117/8)

So the area of AEXD = area of triangle EXD + area of triangle of triangle ADE =

(39/8 + 117/8) = (156/8) = 19.5

So the ratios of BXC / AEXD = 19.5 /19.5  =  1