AoPS - Problem

Source: 67 Polish MO 2016 Second Round - Problem 2

Tags: geometry, circumcircle, tangent, bisector, Poland

mruczek

30.04.2018 23:30

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 $BC$.

Fumiko

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.

jayme

01.05.2018 12:35

Dear, how do you prove PA = PD? Sincerely Jean-Louis

ManuelKahayon

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.

jayme

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

ywq233

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) $.

mathenthusiastic

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 .$\blacksquare$

AlastorMoody

13.03.2019 18:52

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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 $\odot (DEF)$

Kgd

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

Rahmanhabibi

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

Mahdi_Mashayekhi

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 $DEF$.