# Triangle inscribed in another triangle with same proportion on the sides

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.

• "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.
– Nij
Feb 20, 2022 at 9:10

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