90-A generalization Napoleon's theorem associated with the Kiepert hyperbola

Very nice theorem: http://tube.geogebra.org/student/m542855

Let ABC be a triangle, $F$ be the first (or secon) Fermat point, let $K$ be the point on the Kiepert hyperbola. Let $P$ be the point on line FK. The line through P and perpendicular to $BC$ meet $AK$ at $A_0$. Define $A_0,B_0,C_0$ cyclically. Show that $A_0B_0C_0$ is an equilateral triangle. This triangle homothety to the outer(or inner) Napoleon triangle.

Lemma 1:(USA TST 2006, Problem 6) Let $ABC$ be a triangle. Triangles $PAB$ and $QAC$ are constructed outside of triangle $ABC$ such that $AP = AB$ and $AQ = AC$ and $\angle{BAP}= \angle{CAQ}$. Segments $BQ$ and $CP$ meet at $R$. Let $O$ be the circumcenter of triangle $BCR$. Prove that $AO \perp PQ.$

Telv Cohl's proof:

Let $ O' $ be the circumcenter of $ \triangle ACQ $ . Let $ M, N $ be the midpoint of $ CQ, CR $, respectively .

Easy to see $ R \in (O') $ .

Since $ O', M, N, C $ are concyclic , so we get $ \angle AO'O=\angle QCP  $ . ... $ (1) $ Since $ \angle RO'O=\angle BQC, \angle O'OR=\angle CBQ $ ,
so we get $ \triangle ORO' \sim \triangle BCQ $ , hence $ \frac{O'A}{CQ}=\frac{O'R}{CQ}=\frac{O'O}{QB}=\frac{O'O}{CP} $ . ... $ (2) $

From $ (1) $ and $ (2) $ we get $ \triangle AOO' \sim \triangle QPC $ , so from $ OO' \perp PC $ and $ AO' \perp QC \Longrightarrow AO \perp QP $ .

Lemma 2:

Let $ D $ be a point out of $ \triangle ABC $ satisfy $ \angle DBC=\angle DCB=\theta $ .
Let $ E $ be a point out of $ \triangle ABC $ satisfy $ \angle EAC=\angle ECA=90^{\circ}-\theta $ .
Let $ F $ be a point out of $ \triangle ABC $ satisfy $ \angle FAB=\angle FBA=90^{\circ}-\theta $ . Then $ AD \perp EF $ .

Proof of the lemma:

Let $ B' \in AF, C' \in AE $ satisfy $ AB=AB', AC=AC' $ and $ T=BC' \cap CB' $ .

Easy to see $ \triangle ABB' \cup F \sim \triangle ACC' \cup E \Longrightarrow EF \parallel B'C' $ .

From $ \triangle AB'C \sim \triangle ABC' \Longrightarrow  \angle BTC=180^{\circ}-(90^{\circ}-\theta )=90^{\circ}+\theta $ , so combine with $ \angle DBC=\angle DCB=\theta $ we get $ D $ is the circumcenter of $ \triangle BTC $ , hence from lemma 1, we get $ AD \perp B'C' $ . i.e. $ AD \perp EF $

From the lemma we get the following property about Kiepert triangle :
The pedal triangle of the isogonal conjugate of $ K_{90-\phi} $ WRT $ \triangle ABC $ and the Kiepert triangle with angle $ \phi $ are homothetic .
( Moreover, the homothety center of these two triangles is the Symmedian point of $ \triangle ABC $ ! ) (1)

Let $ H_b, H_c $ be the orthocenter of $ \triangle FCA, \triangle FAB $, respectively . ( $ H_b, H_c $ also lie on the Kiepert hyperbola of $ \triangle ABC $ )

Easy to see all $ \triangle A_0B_0C_0 $ are homothetic with center $ K $ , so it is suffices to prove the case when $ P $ coincide with $ F $ .

From Pascal theorem (for $ CKBH_cFH_b $) we get $ AF \perp B_0C_0 $ . Similarly, we can prove $ BF \perp C_0A_0 $ and $ CF \perp A_0B_0 $ , so $ \triangle A_0B_0C_0 $ and the pedal triangle of the isogonal conjugate of $ F $ WRT $ \triangle ABC $ are homothetic , hence from (1) we get $ \triangle A_0B_0C_0 $ and the outer (or inner) Napoleon triangle are homothetic .

Reference:

[1] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&t=622242

[2] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&t=621954

[3] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=148830