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$BB'$ and $CC'$ are altitudes of $\triangle ABC$. $BD$ and $CD$ are tangents of the circumscribed of $\triangle ABC$. $DD' \perp BC$ at $D'$. $AD \cap BC = \{E\}$ and $AD' \cap B'C' = \{E'\}$. Prove that $EE' \perp BC$.

I tried $BB' \cap CC' = \{H\}$ and prove that $AH \parallel EE'$, although I haven't known if there are any ways possible.

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Firstly, it's clear that $D'$ is midpoint of segment $BC$. Then, note that $D'B'=D'C'=D'B=D'C$ (points $B,C,B',C'$ lie on the circle with diameter $BC$). From equalities $\angle B'BC'=90^{\circ}-\angle A$ and $\angle B'D'C'=2\angle B'BC'$ we obtain $\angle B'D'C'=180^{\circ}-2\angle A$. Therefore, (from $D'B'=D'C'$) we get $\angle D'B'C'=\angle D'C'B'=\angle A=\angle B'AC'$. Hence, lines $D'B'$ and $D'C'$ are tangents to circumcircle of triangle $AB'C'$. Now, note that triangles $AB'C'$ and $ABC$ are similar, so points $D'$ and $D$, respectively, are corresponding each other in these triangles. Also $E=AD\cap BC$ and $E'=AD'\cap B'C'$, so construction $(A,B,C,D,E)$ is similar to $(A,B',C',D',E')$. It means that $\frac{AE'}{AD'}=\frac{AE}{AD}$. The last equality implies that $EE'\parallel DD'$. But $DD'\perp BC$, so $EE'\perp BC$, as desired.

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