In acute triangle $ABC$ bisector of angle $BAC$ intersects side $BC$ in point $D$. Bisector of line segment $AD$ intersects circumcircle of triangle $ABC$ in points $E$ and $F$. Show that circumcircle of triangle $DEF$ is tangent to line $BC$.
Problem
Source: 67 Polish MO 2016 Second Round - Problem 2
Tags: geometry, circumcircle, tangent, bisector, Poland
30.04.2018 23:51
Let tangents at $A$ to $\odot (ABC)$ meet $BC$ at $P$ since $PA=PD$ by pop we have $PD^2=PA^2=PE.PF$ so done.
01.05.2018 12:35
Dear, how do you prove PA = PD? Sincerely Jean-Louis
01.05.2018 14:11
Well, you consider the \(A\)-appolonius circle of \(ABC\) and notice that \(P\) must lie on the perpendicular bisector of \(AD\), since the intersection of the tangent to \(A\) at \(ABC\) and \(BC\) is the center of the \(A\)-appolonius circle and \(P\) is the center of the \(A\)-appolonius circle.
01.05.2018 14:21
Dear, thank for this way... Rarely, I am thinking to an angle chasing...can you investigate this way? Sincerely Jean-Louis
02.05.2018 17:31
Let $ EF \cap AD = X , EF \cap BC = S $.Let $ AP \perp AD , AP \cap BC = P $ .Since $ XF \perp AD $, we have $ XS // AP $.So $ PS = SD $ by $ ( P , B , D , C ) = -1 $,then $ SD^2 = SB \cdot SC = SE \cdot SF $ gives $ BC $ tangent to $ \odot (DEF) $.
18.02.2019 20:27
my solution: let $ED,FD\cap \odot (ABC)$ in $M,N$ .since $\angle FEM=\angle FNM$ it is enough to prove $BC\parallel MN$ which is equivalent to prove $\angle BAN=\angle MAC$. simple angle chasing leads $NF,ME$ are bisectors in $\triangle ANM$ so $AD$ also bisect $\angle NAM$ therefore, $\angle BAN=\angle MAC$ as desired .$\blacksquare$
13.03.2019 18:52
Let $AD \cap EF=X$, $AD \cap \odot (ABC)=M_A$, $ED \cap \odot (ABC)=E'$ and $FD \cap \odot (ABC)=F'$. Now observe, $A$ is the reflection of $D$ over $EF$, hence, $D$ is the orthocenter of $\Delta EFM_A$ and since, $\Delta AF'E'$ is the circum-orthic triangle WRT $\Delta EFM_A$, hence, $D$ is the incenter WRT $\Delta AF'E'$ $\implies$ $AF',AE'$ are isogonal WRT $\angle BAC$ hence, $E'F'||BC$ $\implies$ $\angle EFD$ $=$ $\angle EE'F'$ $=$ $\angle EDB$, hence, $BC$ is tangent to $\odot (DEF)$
15.12.2019 21:43
mruczek wrote: In acute triangle $ABC$ bisector of angle $BAC$ intersects side $BC$ in point $D$. Bisector of line segment $AD$ intersects circumcircle of triangle $ABC$ in points $E$ and $F$. Show that circumcircle of triangle $DEF$ is tangent to line $BC$. Why should p be on EF
17.09.2021 03:24
We know that $\angle EFD$ = $\angle AFE$ and $\angle EAD$ =$\angle EDA$ =1/2. $\angle BAC$ + $\angle EAB$ and $\angle ADB$ =1/2. $\angle BAC$ + $\angle ACB$ now if we look at last 2 equations we notice that $\angle EDB$=$\angle ACB$ - $\angle EAB$ = $\angle AFE$ and now proof is completed
16.02.2022 08:26
Let Tangent at $A$ to circumcircle meet $BC$ at $P$. $\angle DAP = \angle BAP + \angle DAB = \angle ACD + \angle DAC = \angle ADP$ so $ADP$ is isosceles so $P$ lies on $EF$ so $PA^2 = PE.PF = PD^2 $ so $PD$ is tangent to $DEF$.