# Prove that $DD' \parallel EE'$.

$$BB'$$ 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.

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 $$EE'||DD'.$$
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.
• Nope both triangles $ABC,AB'C'$ aren't similar. – Love Invariants Mar 22 at 18:03
• Note that $BB'$ and $CC'$ are altitudes of triangle $ABC$. – richrow Mar 23 at 5:36
• No. Lines $B'C'$ isn't parallel to $BC$ so how can those 2 triangles be similar. – Love Invariants Mar 24 at 10:30
• Triangles $ABC$ and $AB'C'$ are similar (vertices in this order). Namely, $\angle AB'C'=\angle ABC$ and $\angle AC'B'=\angle ACB$ because $BB'$ and $CC'$ are altitudes in triangle $ABC$ and quadrilateral $BCB'C'$ is cyclic. – richrow Mar 24 at 11:11