$BB'$ and $CC'$ are altitude of $\triangle ABC$. Point $D'$ is outside $\triangle ABC$ such that $D'B \perp AB$ at $B$ and $D'C \perp AC$ at $C$. $AD \cap B'C' = \{E\}$ and $AD' \cap BC = \{F\}$. Prove that $DD' \parallel EE'$.
I tried using intercept theorem $\left(\dfrac{AE}{AD} = \dfrac{AE'}{AD'}\right)$but I don't know how.
On
Consider triangles $ABC$ and $AB'C'$. Note that they are similar. Indeed, quadrilateral $BCB'C'$ is cyclic (because $\angle BB'C=\angle BC'C=90^{\circ}$), so $\angle AB'C'=\angle ABC$ and $\angle AC'B'=\angle ACB$. Hence, triangles $ABC$ and $AB'C'$ are similar.
Moreover, in this similar triangles point $D$ for triangle $ABC$ corresponds to point $D'$ for triangle $AB'C'$ (because $D$ is point opposite to $A$ on circumcircle $(ABC)$; simlilar for $D'$). Since $E=AD\cap BC$ and $E'=AD'\cap B'C'$ in this triangles we obtain that points $E$ and $E'$ are corresponding to each other in these triangles. Therefore, constructions $(A,B,C,D,E)$ and $(A,B',C',D',E')$ are similar. Hence, $\frac{AE}{AD}=\frac{AE'}{AD'}$, so $EE'$ and $DD'$ are parallel, as desired.
Quadrilaterals $AC'DB'$, $BC'B'C$ and $ACD'B$ are cyclic.
Thus, $$\measuredangle D'AC=\measuredangle D'BC=\measuredangle C'CB=\measuredangle C'B'B=\measuredangle C'AD$$ and $$\measuredangle BCA=180^{\circ}-\measuredangle BC'B'=\measuredangle AC'E.$$ Thus, $\Delta C'AE\sim\Delta CAE'$ and $\Delta DC'A\sim\Delta DCA,$ which gives $$\frac{AE}{AE'}=\frac{AC'}{AC}=\frac{AD}{AD'}$$ and from here $$\frac{AE}{AD}=\frac{AE'}{AD'},$$ which gives $$EE'||DD'.$$