Let $\Delta ABC$ be acute and point $D$ be foot of the altitude from the vertex $C$. Let $E\in\overline{AC}$ s.t. $DE\perp AC$. Let $F,G,H$ belong to $\overline{DC},\overline{DA}$ and $\overline{ED}$ respectively. s.t.: $$|DF|=\frac{1}{3}|DC|,|DG|=\frac{1}{3}|DA|,|EH|=\frac{1}{3}|ED|.$$
Prove $\Delta FGH\sim\Delta ACD$ (not necessarily in that vertex order).
My approach:
Applying the intercept theorem, we obtain the following:
Let $\overrightarrow{DJ}=\frac{2}{3}\overrightarrow{DC}$. Then: $$AC\parallel HJ\parallel GF$$ Let $I\equiv GH\cap DE$.
$I$ is the midpoint of $\overline{HD}$.
Since $\overline{GF}\perp\overline{HD}$ , $\Delta DFH\;\&\;\Delta HGD$ and $\Delta HGI\;\&\;\Delta IGD$ are isosceles. $(\;|HF|=|DF|\;\&\;|HG|=|DG|\;)$ $\implies\Delta HGF\overset{\sim}{=}\Delta GDF$. $$\Delta GDF\sim\Delta ACD\implies\Delta HGF\sim\Delta ACD\;\;\blacksquare$$
Picture:
Question: Apart from methods involving vectors, are there any other non-routine that might include compositions of linear transformations or something totally different worth more focus? Thank you in advance!
As you have demonstrated, we have $|HG|=|DG|$. By Talete's theorem we have: $$\overline{DG}=\overline{GT}=\overline{TA}=\frac{1}{3}\overline{AD}$$ where $T$ is the point of intersection of the line passing throught $J$ and $H$ with $AB$. Also, we have that: $$\overline{CJ}=\overline{JF}=\overline{FD}=\frac{1}{3}\overline{CD}$$ This two results aree very important, also, as you noted before: $$\overline{HG}=\overline{GD}$$ and $$\overline{HF}=\overline{FD}$$ The sides $\overline{FH}, \overline{HG}, \overline{ED}, \overline{EC}$ are in proportion: $$\frac{\overline{FH}}{\overline{CD}}=\frac{\overline{HG}}{\overline{AD}}=\frac{1}{3}$$ Now, we have to focus on the triangles $FGD, JTD, CAD$. This triangles are similar, in fact all have in common the angle $CDA$ and, because what you have said in your post and what I have written here. So, we have: $$\frac{\overline{FG}}{\overline{AC}}=\frac{1}{3}$$ But now we have two triangles that have all sides in proportion, in fact: $$\frac{\overline{HG}}{\overline{AD}}=\frac{\overline{HF}}{\overline{CD}}=\frac{\overline{FG}}{\overline{AC}}=\frac{1}{3}$$ So, the triangles $ACD$ and $FGH$ are similar.