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!

  • $\begingroup$ "the angles don't seem to be the same" - what the angles look like is irrelevant to the mathematical fact. If you can prove similarity, they are similar. $\endgroup$
    – Nij
    Feb 20, 2022 at 9:10

3 Answers 3


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 $x=2-\sqrt{3}$.


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 $x$.


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: $$x=2\pm\sqrt{3}.$$


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