Let $ABC$ be a acute triangle where $\angle BAC =60$. Prove that if the Euler's line of $ABC$ intersects $AB,AC$ in $D,E$, then $ADE$ is equilateral.
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Tags: geometry
25.04.2016 19:08
It is easy to see that $AH=AO$, and that's it
30.04.2016 23:37
abnunc wrote: It is easy to see that $AH=AO$, and that's it PLs. explain why AH=AO. Thank's..
30.04.2016 23:55
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.
01.05.2016 06:23
or $AH=2R\cos{A}=R=AO$
28.06.2016 18:49
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.
29.06.2016 02:56
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.
25.05.2018 15:48
You could also use the Lemma $AH=BC \cdot cotA$(Which isn't hard to prove and easily show that $AO=BO=CO=\frac{BC}{\sqrt3}$
25.05.2018 16:19
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. $\blacksquare$
11.07.2018 14:17
13.11.2019 15:35
#42 at https://artofproblemsolving.com/community/u453822h587522p13472389