How would you solve this geometry problem? (MMC $2015$)

Question

Question: Find the area of the shaded region given $EB=2,CD=3,BC=10$ and $\angle EBC=\angle BCD=90^{\circ}$.

I first dropped an altitude from $A$ to $BC$ forming two cases of similar triangles. Let the point where the altitude meets $BC$ be $X$. Thus, we have$$\triangle BAX\sim\triangle BDC\\\triangle CAX\sim\triangle CEB$$ Using the proportions, we get$$\frac {BA}{BD}=\frac {AX}{CD}=\frac {BX}{BC}\\\frac {CA}{CE}=\frac {AX}{EB}=\frac {CX}{CB}$$ But I'm not too sure what to do next from here. I feel like I'm very close, but I just can't figure out $AX$.

Answer 1

HINT

$\frac {BA}{BD} = \frac {BE}{BE + CD}$ because $\triangle AEB\sim\triangle ACD$

Answer 2

Hint: From the given info, you can compute the sum of the areas of triangles $\triangle EBC$ and $\triangle BDC$: $$ \frac{1}{2}(2\cdot 10)+\frac{1}{2}(3\cdot 10)=25. $$ With a quick observation, you can also compute the sum of the areas of triangles $\triangle EBA$ and $\triangle ACD$: $$ \frac{1}{2}(2\cdot4)+\frac{1}{2}(3\cdot6)=13. $$ Two times the answer you seek is the difference between these 2 sums.

Answer 3

Coordinates:

B (0,0)

E(0,2)

C (10, 0)

D( 10, 3)

BD 10y - 3x = 0

EC x+5y = 10

Find y coordinate of A: 10y - 3x = 0, 3x+15y = 30 => 25y = 30 => y = 6/5

Area size is 10*(6/5)/2 = 6;

Answer 4

You are almost there. Just add the right hand pairs of equations you have so you get $$\frac{AX}{CD}+\frac{AX}{EB}=1$$

Substituting the values, you get $AX=\frac 65$ so the required area is...?