For reference: In the figure below, $ABC$ is a triangle of heights $AD$, $BE$ and $CF$ respectively. Calculate $DF$ knowing that $BD = 3 cm$, $DC = 8 cm$ and $DE = 6 cm$. $CF$, $BE$ e $AD$ are perpendicular.(Answer:$4$)
My progress: $\triangle DEF$ $ is a orthic triangle
Similarities I found $\triangle ABD \sim CBF\sim CHD\\ \triangle AEH \sim \triangle BDH \sim\triangle BEC\\ \triangle AEB \sim AFC\\ \frac{CH}{AB}=\frac{HD}{3}=\frac{8}{AD}\\ \frac{11}{AB}=\frac{BF}{3}=\frac{CE}{AD}\\ \frac{11}{CH}=\frac{BF}{DH}=\frac{CF}{8}\\ \frac{11}{BH}=\frac{EC}{HD}=\frac{BE}{3}\\ $
but can't find the relationship with DF...
Let O be the intersection of the heights. Let $\angle OBC=\alpha$ and $\angle OCD = \theta$. With the properties of a cyclic quadrilateral, you can find the angles marked in the diagram below in terms of $\alpha$ and $\theta$
Now observe $\triangle FDC$ and $\triangle BDE$ They're similar!.
With triangle similarity, you can get the following relationship.
$\frac{x}{3}=\frac{8}{6}$.
By simplifying,
$\boxed{x=4}$