# Prove that $EE' \perp BC$.

$$BB'$$ 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.

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.