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Concurrence of angle bisectors
proglote   65
N 4 minutes ago by smbellanki
Source: Brazil MO #5
Let $ABC$ be an acute triangle and $H$ is orthocenter. Let $D$ be the intersection of $BH$ and $AC$ and $E$ be the intersection of $CH$ and $AB$. The circumcircle of $ADE$ cuts the circumcircle of $ABC$ at $F \neq A$. Prove that the angle bisectors of $\angle BFC$ and $\angle BHC$ concur at a point on $BC.$
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proglote
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smbellanki
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rmo 2016 odisha region,india
amc789   7
N Jul 24, 2017 by Kaling21
Source: RMO 2016 ODISHA REGION
Let $ABC$ be the triangle,$AD$ is an altitude and $AE$ is a median. Assume $B$,$D$,$E,C$ lie in that order on the line $BC$ . Suppose the in center of the triangle $ABE$ lie on $AD$ and the incentre of triangle $ADC$ lies on $AE$. Find with proof,the angles of triangle $ABC$.
7 replies
amc789
Oct 23, 2016
Kaling21
Jul 24, 2017
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rmo 2016 odisha region,india
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Source: RMO 2016 ODISHA REGION
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amc789
45 posts
amc789
#1 h  Oct 23, 2016, 2:25 PM • 1 Y
Y by Adventure10
Let $ABC$ be the triangle,$AD$ is an altitude and $AE$ is a median. Assume $B$,$D$,$E,C$ lie in that order on the line $BC$ . Suppose the in center of the triangle $ABE$ lie on $AD$ and the incentre of triangle $ADC$ lies on $AE$. Find with proof,the angles of triangle $ABC$.
This post has been edited 1 time. Last edited by lehungvietbao, Jun 26, 2018, 5:40 AM
Reason: LaTeX
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Kaling21
35 posts
Kaling21
#2 h  May 26, 2017, 4:03 AM • 2 Y
Y by Adventure10, Mango247
Give the correct question.couldnt understand
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Kaling21
35 posts
Kaling21
#3 h  May 26, 2017, 4:04 AM • 2 Y
Y by Adventure10, Mango247
It's 60°,90° and 30°
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eisirrational
1113 posts
eisirrational
#4 h  May 26, 2017, 4:05 AM • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be the triangle,$AD$ is an altitude and $AE$ is a median. Assume $B$,$D$,$E$,$C$ lie in that order on the line $BC$ . Suppose the in center of the triangle $ABE$ lie on $AD$ and the incentre of triangle $ADC$ lies on $AE$. Find with proof,the angles of triangle $ABC$.
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Virgil Nicula
7054 posts
Virgil Nicula
#5 h  May 26, 2017, 7:18 PM • 1 Y
Y by Adventure10
See the proposed problem P7 from here
This post has been edited 9 times. Last edited by Virgil Nicula, May 26, 2017, 7:40 PM
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Kaling21
35 posts
Kaling21
#7 h  May 27, 2017, 12:41 PM • 2 Y
Y by Adventure10, Mango247
http://imgur.com/a/uZD4W .here's the solution
Later you can put the value of theta and find the angles
This post has been edited 1 time. Last edited by Kaling21, May 27, 2017, 12:43 PM
Reason: Incomplete
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RC.
439 posts
RC.
#8 h  Jun 22, 2017, 3:22 PM • 1 Y
Y by Adventure10
@Kalinga21 Really needs to be that long. Here is a fairly shorter one. Clearly, \(ED = BD = 1/2 BE = 1/2 EC => \frac{ED}{CE} = \frac{1}{2} = \frac{AD}{AC}\) [[ Angle Bisector theorem]] \(=> \angle ACD = 30^{\circ} , \angle ABC = 60^{\circ}\) You can find \(\angle BAC =90^{\circ}\) As far as my calculating skills go.
\(K.I.P.K.IG.\)
This post has been edited 1 time. Last edited by RC., Jun 22, 2017, 3:27 PM
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Kaling21
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Kaling21
#9 h  Jul 24, 2017, 4:46 PM • 2 Y
Y by Adventure10, Mango247
Yeah RC..probably you are right
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