The line $l$ intersects the extension of $AB$ in $D$ ($D$ is nearer to $B$ than $A$) and the extension of $AC$ in $E$ ($E$ is nearer to $C$ than $A$) of triangle $ABC$. Suppose that reflection of line $l$ to perpendicular bisector of side $BC$ intersects the mentioned extensions in $D'$ and $E'$ respectively. Prove that if $BD+CE=DE$, then $BD'+CE'=D'E'$.
Problem
Source: Iran second round 2011
Tags: geometry, geometric transformation, reflection, circumcircle, incenter, perpendicular bisector, geometry proposed
goldeneagle
29.04.2011 12:29
let T be the midpoint of arc BC in circumcircle of triangle ABC. let K be a point on DE such that DK=DB. so EK=EC. now prove that BICK(I is incentre of ABC) is cyclic and T is the center of circle (because TB=TC=TI). so DT is bisector of BDE.(because TB=TK SO T is on perpendicular bisector of BK and DKB is isosceles so DT is bisector of BDE) now let P be the intersection point of DE and the perpendicular bisector of BC and we know that PT is bisector of DPE'. now Prove that there is a circle inscribed ADPE' and this result give the solution easily.
ArefS
29.04.2011 12:32
It is enough to consider the incircle of $\triangle ADE$
mahanmath
29.04.2011 14:14
goldeneagle wrote: let T be the midpoint of arc BC in circumcircle of triangle ABC. let K be a point on DE such that DK=DB. so EK=EC. now prove that BICK(I is incentre of ABC) is cyclic and T is the center of circle (because TB=TC=TI). so DT is bisector of BDE.(because TB=TK SO T is on perpendicular bisector of BK and DKB is isosceles so DT is bisector of BDE) now let P be the intersection point of DE and the perpendicular bisector of BC and we know that PT is bisector of DPE'. now Prove that there is a circle inscribed ADPE' and this result give the solution easily. Exactly mine! It`s a really nice fact , but I think it should not originaly for Iran !
darrande
17.05.2011 15:57
OK, I HAVE REMOVED IT.
amirreza_math
05.08.2012 14:25
Consider the reflections of AB and AC asA'B and A'C.the line through D'E' meet AB,AC atD'',E'' respectively.consider point m on line D''E''such that MD'' =BD''then ME''=CE".M' is a point on D'E'such thatBCMM' is cyclic.by some angle chasing we arrive to M'D'=BD',M'E'=CE' and so we are done