# $\sqrt{A(ABCD)} =\sqrt{A(ABE)}+ \sqrt{A(CDE)}$

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.

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Which book is this from? Seems interesting. – Sawarnik Feb 8 '14 at 11:03
@Sawarnik: It is written in Portuguese. – spohreis Feb 11 '14 at 11:22

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:

$$$\dfrac{[AED]}{[BEC]} + 1 = 2 \sqrt{\dfrac{[ABE]}{[BEC]} \dfrac{[CDE]}{[BEC]}}$$$. (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.

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I apologize for how rough the proof is - I will clean it up in a bit. – Michael Zhao Oct 4 '12 at 19:34
There is no need to clean it up. It is perfect! – spohreis Oct 4 '12 at 22:15