Prove that $DD' \parallel EE'$.

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 $$EE'||DD'.$$

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