$\sqrt{A(ABCD)} =\sqrt{A(ABE)}+ \sqrt{A(CDE)}$ I am self-studying Euclidean Geometry, and I want to solve the following exercise.
Let $ABCD$ a trapezoid with bases $AB, CD$, and $AC, BD$ are the lateral sides. If $E$ is the point of intersection of its diagonals, then we have $$\sqrt{A(ABCD)} =\sqrt{A(ABE)}+ \sqrt{A(CDE)}$$  where $A(ABCD)$ denotes de area of the trapezoid $ABCD$, $A(ABE)$ denotes the area of the triangle $ABE$ and so on.
I was not able to  prove that. I would appreciate any help.
 A: Let me switch the notation to brackets: that is let $A(ABCD) = [ABCD]$. Then what we want to prove is
$\sqrt{[ABCD]} = \sqrt{[ABE]} + \sqrt{[CDE]}.$
Squaring both sides and subtracting $[ABE]$ and $[CDE]$ from both sides gives:
$[AED] + [BEC] = 2\sqrt{[ABE][CDE]}$. 
Dividing both sides by $[BEC]$ gives:
$\begin{equation} \dfrac{[AED]}{[BEC]} + 1 = 2 \sqrt{\dfrac{[ABE]}{[BEC]} \dfrac{[CDE]}{[BEC]}} \end{equation}$. (1)
We prove this instead (if this is true we can go back to the original equation given).
Proof. Since triangles $ABE$ and $BEC$ have the same height from $B$, the ratio of their areas is given by $AE/EC$. Similarly, $[CDE]/[BEC] = ED/BE$. But observe that by AA similarity, $ABE$ and $CDE$ are similar, and hence $BE/ED = AE/EC$. This means $\dfrac{AE}{EC} \cdot \dfrac{ED}{BE} = 1$, and so $\dfrac{[ABE]}{[BEC]} \dfrac{[CDE]}{[BEC]} = 1$. Thus the RHS of (1) equals 2.
Notice further that $[ABD] = [ABC]$ and subtracting $[ABE]$ from both sides gives
$[ABD]-[ABE] = [ABC]-[ABE] \Leftrightarrow [AED] = [BEC]$,
and so the LHS of (1) is: $\dfrac{[AED]}{[BEC]} + 1 = 2$. QED.
