Let ABC be a acute triangle where ∠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=2RcosA=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 ∡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 ∠ OAE = ∠ HAD and HA = 2OM ( theorem). Then ∠ MOC = 60∘ so OA = HA. Then △ DHA ≅ △ AOE ⟶ △ DAE is equilateral.
25.05.2018 15:48
You could also use the Lemma AH=BC⋅cotA(Which isn't hard to prove and easily show that AO=BO=CO=BC√3
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 ∠BAC. So, O also lies on D′E′, which leads to the conclusion. ◼. 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∥OH. Now the conclusion is obvious. ◼
11.07.2018 14:17
13.11.2019 15:35
#42 at https://artofproblemsolving.com/community/u453822h587522p13472389