Concurrency involving equilateral triangles constructed outside a triangle Let $ABC$ be a triangle. On the exterior of sides $AB$ and $AC$, construct two equilateral triangles $ABE$ and $ACF$. Let $H$ be the orthocenter of $ABC$ and $P$, $Q$ be the centers of $ABE, ACF$ respectively. I found out that $AH, EQ, FP$ are concurrent, and I remembered that it was a theorem in school geometry, but I couldn't prove it or find out its origin.
I also found (by GeoGebra) that: If $M, N$ are the midpoints of $AC, AB$ respectively then both $MP, NQ, AH$ and $EM, FN, AH$ are concurrent too, but I haven't proven it. Is this a well-known result in plane geometry?

Please give me some hints or prove/disprove these results for me. Thanks
 A: This is a very nice problem!
Lemma
Consider the following diagram, where we construct right triangles externally to an arbitrary triangle $\bigtriangleup ABC$. If $\;\measuredangle ABD=\measuredangle CBE \;$ and $\; BG \perp AC$; $$DC, BG, EA$$ are concurrent.

Proof.
Apply the law of sines on $\bigtriangleup CAD$ and $\bigtriangleup CBD$:
$$\frac{CD}{\sin(90°+\alpha+\beta)}=\frac{AD}{\sin \zeta} \tag{1}$$
$$\frac{CD}{\sin(\measuredangle ABD+\gamma+\delta)}=\frac{AD/(\sin\measuredangle ABD)}{\sin \epsilon} \tag{2}$$
Dividing $(1)$ by $(2)$:
$$\frac{\sin(\measuredangle ABD+\gamma+\delta)}{\sin(90°+\alpha+\beta)}=\frac{\sin \epsilon \cdot (\sin\measuredangle ABD)}{\sin \zeta} \tag{3}$$
Applying the law of sines on $\bigtriangleup ACE$ and $\bigtriangleup ABE$ we also get:
$$\frac{\sin(\measuredangle CBE+\gamma+\delta)}{\sin(90°+\epsilon+\zeta)}=\frac{\sin \beta \cdot (\sin\measuredangle CBE)}{\sin \alpha} \tag{4}$$
Since $\bigtriangleup ABG$ and $\bigtriangleup CBG$ are right triangles:
$$\sin\gamma=\sin(\alpha+\beta) \;\; \text{and} \; \sin\delta=\sin(\epsilon+\zeta)\tag{5}$$
Dividing $(3)$ by $(4)$ and substituting from $(5)$:
$$\frac{\sin(\alpha)}{\sin(\beta)}×\frac{\sin(\gamma)}{\sin(\delta)}×\frac{\sin(\epsilon)}{\sin(\zeta)}= 1$$
Hence, by (Trigonometric) Ceva's theorem, the lines are concurrent. We note that this concurrency is independent of the measures of the angles $\measuredangle ABD$ and $\measuredangle CBE$ as long as they are equal.

Back to the original figure,

By the reverse of Thales' Theorem $NM \parallel BC$, so $AH \perp NM$. Construct the line segments $MP, AP, NQ, AQ\;$ and we get the desired configuration for our lemma (the main triangle being $\bigtriangleup AMN$). So, $MP, NQ, AH$ are concurrent. By similar constructions and using our lemma again both $EM, FN, AH$ and $EQ, FP, AH$ are also concurrent.  $\; \blacksquare$

I suspect there is a simpler way to prove the Lemma (might be related to Jacobi's Theorem).
A: Welcome to MSE! This is vaguely related to the Fermat Point. Hope this helps.
A: Well, this should not be consider as an answer, but as a complement for the accepted answer of dodoturkoz, a more elementary proof for the lemma.

Let $K$ be the intersection of $BG$ and the perpendicular line of $AE$ passing through $C$.
Then, $\widehat{KCB}=\widehat{AEC}$.
Also $\widehat{KBC}=\dfrac{\pi}{2}+\widehat{ACB}=\widehat{BCE}+\widehat{ACB}=\widehat{ECA}$.
Thus the two triangles $KBC$ and $ACE$ are similar therefore: $\dfrac{KB}{AC}=\dfrac{BC}{CE}$.
On the other hand, $\Delta BCE\backsim \Delta BAD$, hence,
$\dfrac{BC}{CE}=\dfrac{AB}{AD}$.
Thus $\dfrac{KB}{AC}=\dfrac{AB}{AD}$.
Moreover: $\widehat{KBA}=\dfrac{\pi}{2}+\widehat{BAC}=\widehat{BAD}+\widehat{BAC}=\widehat{DAC}$.
Therefore $\Delta KBA\backsim \Delta CAD$, and then $\widehat{BAK}=\widehat{ADC}$. Hence $CD\perp AK$.
As so, let $F$ be the intersection of $CD$ and $AE$, $F$ should be the orthocenter of $KAC$ and therefore $F$ is on $BG$.
Proof for the lemma is completed.
