It is easy to see that $AH=AO$, and that's it

abnunc wrote: It is easy to see that $AH=AO$, and that's it PLs. explain why AH=AO. Thank's..

MathGan wrote: abnunc wrote: It is easy to see that $AH=AO$, and that's it PLs. explain why AH=AO. Thank's.. Use the Incenter-Excenter lemma.

or $AH=2R\cos{A}=R=AO$

MathGan wrote: abnunc wrote: It is easy to see that $AH=AO$, and that's it PLs. explain why AH=AO. Thank's.. Just see that $BHOC$ is cylcic (which is easy using that $\measuredangle BAC=60$) and then it´s trivial by angle chasing.

Let $M$ midpoint of $BC$, $H$ orthocenter and $O$ circumcenter of $(ABC)$. It´s easy to see that $\angle$ $OAE$ = $\angle$ $HAD$ and $HA$ = $2OM$ ( theorem). Then $\angle$ $MOC$ = $60^\circ$ so $OA$ = $HA$. Then $\triangle$ $DHA$ $\cong$ $\triangle$ $AOE$ $\longrightarrow$ $\triangle$ $DAE$ is equilateral.

You could also use the Lemma $AH=BC \cdot cotA$(Which isn't hard to prove and easily show that $AO=BO=CO=\frac{BC}{\sqrt3}$

Let $O$ be the circumcenter and $H$ be the orthocenter. Through $H$, draw a line intersecting $AB,AC$ at $D',E'$, such that $AD'E'$ is equilateral. Notice that $AH=AO$ and that $AH,AO$ are isogonal WRT $\angle BAC$. So, $O$ also lies on $D'E'$, which leads to the conclusion. $\blacksquare$. Here's another Solution: Let the circle with center $A$ and radius $AO$ (also passing through $H$) meet $AB,AC$ at $X,Y$. See that $AXY$ is equilateral and that chords $XH,YO$ subtend equal angles at the center $A$. So, $XY\parallel OH$. Now the conclusion is obvious. $\blacksquare$

#42 at https://artofproblemsolving.com/community/u453822h587522p13472389