Let ABC be an acute-angled triangle in which ∠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 ∠AXB and ∠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
07.12.2014 13:52
My solution: DE∥AC⟹ADDB=CEEB⟹AXXB=CYYB ⟹AXXC=CYYA⟹AY=CX⟹sin∠CABsin∠BCA=BCBA=cos∠BCAcos∠CAB ⟹sin2∠CAB=sin2∠BCA⟹∠CAB=∠BCA ( ∵ \triangle ABC is an acute-angled triangle) \Longrightarrow BO \perp AC Q.E.D
07.12.2014 14:23
the same as TelvCohl until AY=CX after note that BXC and BYA are isometric
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
10.12.2014 07:39
aditya21 wrote: also triangle AXC , CYA triangle are similar A-X-C are collinear i think..
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?
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.
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....
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
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 ?
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)
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:


25.09.2024 20:01