Ratio of parts of a triangle

Question

In the diagram above, segment DE is parallel to segment BC and the ratio of the area of triangle AED to the area of trapezoid EDBC is 1:2.

How can I find the ratio of AE to AC?

So far, I got the following:

• $\Delta ADE$ ~ $\Delta ABC$
• $\frac{Area \Delta ADE}{Area \Delta ABC}$ = $\frac{AE^2}{AC^2}$

Answer 1

First of all, find the ratio between the area of $ADE$ and the area of $ABC$:

$$S(ABC)=S(ADE)+S(EDBC)=S(ADE)+2S(ADE)=3S(ADE)$$

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Then, use it in order to find the ratio between the length of $AE$ and the length of $AC$:

$$\frac{S(ADE)}{S(ABC)}=\frac13\implies\frac{AE}{AC}=\sqrt{\frac13}$$

Answer 2

You have basically all the pieces you need, all you need to notice now is that the area of $\triangle ABC$ is the area of trapezoid $EDBC$ plus the area of $\triangle ADE$, so the area of $\triangle ABC$ is $3$ times the area of $\triangle ADE$ which gives you a final ratio of $\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}$.