euclidean-geometrygeometry
$M$ is the orthocenter of $\triangle ABC$. $D$ is a point outside of $\triangle ABC$ such that $\widehat{ABD} = \widehat{DBC} = \widehat{DMC}$. $N$ is the orthocenter of $\triangle BCD$. Let $P$ and $Q$ respectively be the circumcenters of $\triangle ABC$ and $\triangle DAB$. Prove that $NP = NQ$.
I tried to let $R$ be the circumcenter of $\triangle BCD$ and prove that $\triangle BRM = \triangle BNQ$. But I still don't know how.
I believe you know the fact that circle $BCH$ is a reflection of circle $ABC$ acros $BC$. Similary is true for the other two circles.
So $$AC' = AO = CO = CA'$$
and by easy angle chase you can see $AC'||CA'$ so $A'CAC'$ is paralelogram, so $A'C' = AC$ and we are done.
Based on your notations and what you have: $$B,A,E,F,C~{\rm are~concyclic~on~the~circumcircle~}(ABC)~{\rm with~center~}O,$$ $$AB\parallel FC~{\rm and~}AE\parallel BC.$$ Let $H$ be the orthocenter of triangle $ABC$. Let $C'$ be the alternative intersection of $CH$ with $(ABC)$. Let $A'$ be the alternative intersection of $AH$ with $(ABC)$. Then $AA'$ (resp. $CC'$) being orthogonal transversal of parallel lines $AE$ and $BC$ (resp. $AB$ and $FC$), one has $$\angle A'AE=90^\circ~({\rm resp.}~\angle C'CF=90^\circ).$$ It follows that both $A'E$ and $C'F$ are diameters of $(ABC)$, so $A'E\cap C'F=O$. Now the six points $C',A,E,A',C,F$ on circle $(ABC)$ satisfy $$CC'\cap AA'=H,C'F\cap EA'=O,AF\cap EC=D.$$ So by Pascal's theorem, $H,O,D$ are collinear. Since $HO$ coincides with the Euler line $GO$, the same is true for $G,O,D$. QED
Best Answer
Here is another solution which came from "mirror circle"
since the proof is very simple, I don't give details here.
2.
solution:
let circle R' is the mirror circle of circle R, connect $RR',OR'$, extend $BD$ to circle R at point $E$, connect $NE$,$NC$,
it is trivial $\angle NEB= \angle NCB $, $N$ is orthocenter $\implies NC\perp BD \implies \angle NEB + \angle DBC =\dfrac{\pi}{2}=\angle NEB+\angle EBA \implies NE \perp AB $
$\angle NDE= \angle BDF, \angle BDF+\angle DBC =\dfrac{\pi}{2} \implies \angle NDE= \angle NED \implies ND=NE=AM=RR'$
$ AB $ on circle $O$ & circle $R \implies OR \perp AB $ with same reason, $OR'\perp BD, RR'\perp BC \implies \angle ROR'= \angle ABD ,\angle RR'O= \angle DBC \implies RR'=OR=NE ,OR // NE \implies NO=ER$
it is trivial $NR=ER$ (radius) $\implies NO=NR$