Triangle inscribed in another triangle with same proportion on the sides

Question

As shown in the figure below, $\triangle DEF$ is a triangle inside $\triangle ABC$.
Given $$AD:DB=BE:EC=CF:FA=1:x$$
Such that the area of $\triangle ABC$ is two times the area of $\triangle DEF$.
Find $x$

I have no idea here. My first thought is that we could make parallel lines on $D, E, F$, though that would split $\triangle DEF$ into parts, which isn't ideal.

Another idea was similar triangles, due to the fact that the sides are proportional. But the angles don't seem to be the same.

Please help!

Answer 1

Recall that, when two triangles have one angle equal, their areas are proportional to the product of the adjacent sides. This is, for example, embodied in so-called sine formula.

So, as $AD=\frac{1}{1+x}AB$ and $AF=\frac{x}{1+x}AC$, then $A(\triangle ADF)=\frac{x}{(1+x)^2}A(\triangle ABC)$. The areas of $\triangle BED$ and $\triangle CFE$ end up the same, so altogether (taking away the areas of those three triangles from the area of $\triangle ABC$):

$$A(\triangle DEF)=A(\triangle ABC)\left(1-3\frac{x}{(1+x)^2}\right)$$

and now, all you need to do is solve the equation $1-\frac{3x}{(1+x)^2}=\frac{1}{2}$. This has two solutions: $x=2+\sqrt{3}$ or $x=2-\sqrt{3}$.

Answer 2

Hint: Can you show that $AD = \dfrac{AB}{1+x}, ~AF = \dfrac{x \cdot AC}{1+x}$?

So, what is $[ADF]$ in terms of $[ABC]$?

And then show that $~[BDE] = [CFE] = [ADF]$

Finally use $ \displaystyle [BDE] + [CFE] + [ADF] = \frac 12 [ABC]$ to find $x$.

Answer 3

A linear algebra based approach:

Let $\vec{u}=\vec{AB}, \vec{v}=\vec{AC}$. Then the area of $\Delta (ABC)$ is just $$\frac12|\vec{u}\times \vec{v}|$$

We have \begin{eqnarray*} \vec{DF}&=&\vec{DA}+\vec{AF}=\frac{-1}{1+x}\vec{u}+\frac{x}{1+x}\vec{v}\\\\\vec{DE}&=& \vec{DB}+\vec{BE}=\frac{x}{1+x}\vec{u}+\frac{1}{1+x}(\vec{v}-\vec{u})= \frac{x-1}{1+x}\vec{u}+\frac{1}{1+x}\vec{v} \end{eqnarray*}

Thus \begin{eqnarray*}\Delta(DEF)&=&\frac12|\vec{DF}\times\vec{DE}|\\\\&=& \frac12\left|\left(\frac{-1}{1+x}\vec{u}+\frac{x}{1+x}\vec{v}\right)\times\left(\frac{x-1}{1+x}\vec{u}+\frac{1}{1+x}\vec{v} \right)\right|\\&=& \frac12|\vec{u}\times \vec{v}|\left|\frac{x^2-x+1}{(1+x)^2}\right|\\&=& \Delta(ABC)\left|\frac{x^2-x+1}{(1+x)^2}\right| \end{eqnarray*}

Thus we must solve $$\frac{x^2-x+1}{(1+x)^2}=\pm\frac12.$$

Either $$x^2-4x+1=0,\qquad{\rm or}\qquad x^2+1=0.$$

Only the first of these has real solutions: $$x=2\pm\sqrt{3}.$$