Let $ABC$ be an acute-angled triangle in which $\angle ABC$ is the largest angle. Let $O$ be its circumcentre. The perpendicular bisectors of $BC$ and $AB$ meet $AC$ at $X$ and $Y$ respectively. The internal angle bisectors of $\angle AXB$ and $\angle BYC$ meet $AB$ and $BC$ at $D$ and $E$ respectively. Prove that $BO$ is perpendicular to $AC$ if $DE$ is parallel to $AC$.
Problem
Source: Mumbai Region RMO 2014 Problem 3
Tags: trigonometry, geometry, angle bisector
TelvCohl
07.12.2014 13:52
My solution: $ DE \parallel AC \Longrightarrow \frac{AD}{DB}=\frac{CE}{EB} \Longrightarrow \frac{AX}{XB}=\frac{CY}{YB} $ $ \Longrightarrow \frac{AX}{XC}=\frac{CY}{YA} \Longrightarrow AY=CX \Longrightarrow \frac{\sin \angle CAB}{\sin \angle BCA}=\frac{BC}{BA}=\frac{\cos \angle BCA}{\cos \angle CAB} $ $ \Longrightarrow \sin 2 \angle CAB=\sin 2 \angle BCA \Longrightarrow \angle CAB=\angle BCA $ ( $ \because $ $ \triangle ABC $ is an acute-angled triangle) $ \Longrightarrow BO \perp AC $ Q.E.D
PROF65
07.12.2014 14:23
the same as TelvCohl until AY=CX after note that BXC and BYA are isometric
aditya21
09.12.2014 08:08
my proof = as $XD,YE$ are angle bisectors thus we have $(AX/XB) = (AD/DB) $ and $ ( CY/YB) = (CE/EB) $ now $ DE\parallel AC\Longrightarrow\frac{AD}{DB}=\frac{CE}{EB}\Longrightarrow\frac{AX}{XB}=\frac{CY}{YB} = (AX/XC) = (CY/YA) $ now triangle $AXB , CYB$ triangle. are similar thus, $ (AX/CY) = (XB/BY) = ( AB/BC) $ and $ \angle BAX = \angle BCY $ and $ \angle AXB = \angle CYB$ also triangle $ AXC , CYA $ triangle are similar but , $ AC$ is common side opp. to common angle. thus , both triangles are congruent. thus, $ AY = CX \Longrightarrow\ AB = BC $ thus triangle $ BOY $ and $ BOX$ are congruent, thus $BO$ is angle bisector and hence as $ ABC $ is isosceles thus $\Longrightarrow\ BO\perp AC $
Levinay
10.12.2014 07:39
aditya21 wrote: also triangle $ AXC , CYA $ triangle are similar A-X-C are collinear i think..
Levinay
10.12.2014 07:50
PROF65 wrote: the same as TelvCohl until AY=CX after note that BXC and BYA are isometric isometric means isosceles?
aditya21
10.12.2014 09:33
Levinay wrote: aditya21 wrote: also triangle $ AXC , CYA $ triangle are similar A-X-C are collinear i think.. can you show me your diagram. i dont think so still please show your diagram.
Levinay
10.12.2014 10:41
aditya21 wrote: Levinay wrote: aditya21 wrote: also triangle $ AXC , CYA $ triangle are similar A-X-C are collinear i think.. can you show me your diagram. i dont think so still please show your diagram. its there in the problem itself...bisector meets AC at X....
PROF65
10.12.2014 16:22
Levinay wrote: PROF65 wrote: the same as TelvCohl until AY=CX after note that BXC and BYA are isometric isometric means isosceles? means have the same angles and side lengths note that BXY is isoceles
guptaamitu1
10.10.2020 22:40
aditya21 wrote: my proof = as $XD,YE$ are angle bisectors thus we have $(AX/XB) = (AD/DB) $ and $ ( CY/YB) = (CE/EB) $ now $ DE\parallel AC\Longrightarrow\frac{AD}{DB}=\frac{CE}{EB}\Longrightarrow\frac{AX}{XB}=\frac{CY}{YB} = (AX/XC) = (CY/YA) $ now triangle $AXB , CYB$ triangle. are similar thus, $ (AX/CY) = (XB/BY) = ( AB/BC) $ and $ \angle BAX = \angle BCY $ and $ \angle AXB = \angle CYB$ also triangle $ AXC , CYA $ triangle are similar but , $ AC$ is common side opp. to common angle. thus , both triangles are congruent. thus, $ AY = CX \Longrightarrow\ AB = BC $ thus triangle $ BOY $ and $ BOX$ are congruent, thus $BO$ is angle bisector and hence as $ ABC $ is isosceles thus $\Longrightarrow\ BO\perp AC $ can you please explain why triangles AXB, CYB are similar ?
guptaamitu1
10.10.2020 22:42
in my solution, I simply proved that BO is perpendicular to DE (though, I agree that the official solution is more elegant and faster)
NTistrulove
12.01.2022 13:27
Another Solution. Edit: The second attachment is the first part of the solution and the first attachment is the last part..
Attachments:
Jishnu4414l
25.09.2024 20:01
It suffices to show that $\triangle ABC$ is $B$-isosceles.
Clearly, $\angle BED=\angle BCA$. Also, $\angle BXD=\frac{1}{2}\angle AXB=\angle BCA$. This gives us that $\angle BXD=\angle BED\implies$ $B,E,X,D$ are concyclic. Analogously, we get that $B,E,Y,D$ are concyclic. Thus, $X,Y,E,D$ are concyclic. Since $XY\parallel DE$, we must have that $XYDE$ is an isosceles trapezoid. So, $\angle XDE=\angle YED$. Notice that $\angle XDE=\angle DXA=\angle BCA$, and that $\angle YED=\angle EYC=\angle BAC$. Thus, we have that $\angle BAC=\angle BCA$, and we are done!