Let $ABC$ be an isosceles triangle with base $AB$. Line $\ell$ touches its circumcircle at point $B$. Let $CD$ be a perpendicular from $C$ to $\ell$, and $AE$, $BF$ be the altitudes of $ABC$. Prove that $D$, $E$, and $F$ are collinear.
Problem
Source: Sharygin Geometry Olympiad 2014 - Problem 3
Tags: geometry, circumcircle, Kosnita
Pirkuliyev Rovsen
15.11.2014 20:08
See here http://geometry.ru/olimp/2014/zaochsol.pdf
I_m_vanu1996
19.11.2014 05:00
$DB$ is tangent to the circumcircle of $ABC$,so $\angle {CBD}=\angle {CAB}$,also $\angle {BFA}=\angle {CDB}=90$,hence $\angle {BCD}=\angle {FBA}$ ,but $FE$ and $AB$ are parallel,so $\angle {FBA}=\angle {BFE}=\angle {BCD}$ but $CFBD$ is cyclic ,so $\angle {DCB}=\angle {DFB}$,hence $D,E,F$ are collinear.
TelvCohl
19.11.2014 11:10
My solution: Easy to see $ EF \parallel AB $ . ... $ (*) $ Since $ B, C, D, F $ are concyclic , so we get $ \angle BFD=\angle BCD=90-\angle DBC=90-\angle BAC=\angle FBA $ , hence we get $ FD \parallel AB $ . Combine with $ (*) $ we get $ D, E, F $ are collinear . Q.E.D
jayme
19.11.2014 11:21
Dear Mathlinkers, 1. EF // AB 2. B, C, D and F concyclic 3. according to a particular case of the Reim's theorem, FD // AB and we are done. Sincerely Jean-Louis
mathetillica
12.02.2019 05:16
Sorry for bumling this, Solution with Simson line: Let $AE$ meet $CD$ at $P$. Since $\angle CBD=\angle A$ so $\angle BCD=90-\angle A=\angle 90-B$ so $\angle CPA=\angle B$. Which gives $CABP$ is cyclic.But $D,E,F$ are perpendiculars from $B$ to the sides of $\Delta ACP$, hence we are done by Simson's line lemma.
khanhnx
10.03.2019 17:15
We have: $\widehat{CFE}$ = $\widehat{BAC}$ = $\widehat{CBD}$ = $\widehat{CFD}$ then: $D$, $E$, $F$ are collinear
PCChess
16.07.2020 23:13
Construct line $DF$. We will show that $E$ lies on $DF$. Let $N$ be a point on $\ell$ that is on the other side of $B$ as $D$. First, it is easy to see that $EF \parallel BA$. Also note that $CFBD$ is cyclic since $\angle CFB=\angle CDB=90$. Then, $\angle ABN = \angle ACB = \angle FDB$, so $DF \parallel BA$. Since $EF$ and $DF$ are both parallel to $BA, D, E, $ and $F$ must be collinear.
Williamgolly
18.09.2020 22:36
Let AE hit (ABC) at X> Since <CXA=<CBD=<<CAB=<CBA. Now, since CFBD is cyclic, we have <FDB = <FCB. Also, from cyclic quadrilateral EXBD, we have <EDB = <EXB = <ACB. Since <EDB=<FDB, we have that D,E,F are colliinear.
OlympusHero
03.09.2021 01:47
Draw in the final altitude from $C$ to $AB$. Let the intersection point be $G$. I claim that $EDBG$ is a parallelogram. We'll first show that $BD=BG$. It's easy to see $\angle CBD = \angle CAB = \angle CBA$, so $\triangle GBC$ is congruent to $\triangle DBC$, meaning $BD=BG$. Next note that $EG=\frac{AB}{2}=GB=BD$ and $\angle EBG = \angle GEB = \angle EBD$, so $EDBG$ is a parallelogram as desired. This means $DE$ is parallel to $AB$, but $EF$ is also parallel to $AB$, so $D,E,F$ are collinear as desired.
jasperE3
03.09.2021 06:20
Claim 1: $B,C,D,F$ are concyclic Since $\angle BFC=90^\circ=180^\circ-\angle BDC$. Claim 2: $FE\parallel AB$ Let $G$ be the foot of the perpendicular from $C$, and let $H$ be the orthocenter of $\triangle ABC$. Then it is true since: $$\angle FEH=90^\circ-\angle CHE=90^\circ-\angle AHG=\angle EAB.$$ Claim 3: $FD\parallel AB$ $\angle BFD=\angle BCD=90^\circ-\angle CBD=90^\circ-\angle ABC=\angle EAB=\angle ABF$ Then $D,E,F$ are collinear. $\square$
SPHS1234
03.09.2021 06:25
Can someone tell me where the collection of Shargyin papers can be found?
jasperE3
03.09.2021 08:50
SPHS1234 wrote: Can someone tell me where the collection of Shargyin papers can be found? here
BVKRB-
03.09.2021 09:56
Let $AE$ meet $\odot(ABC)$ at $G$ Then $$\angle GAB = \angle GCB = \angle EAB = \frac{\angle BCA}{2} = 90^{\circ} - \angle DBC =\angle DAB \implies A-G-D $$Simson lines on $\triangle AGC$ finishes $\blacksquare$
SPHS1234
03.09.2021 14:42
jasperE3 wrote: SPHS1234 wrote: Can someone tell me where the collection of Shargyin papers can be found? here hey there is some problem arising when you click the pdf sign (2021) Shargyin)..
jasperE3
03.09.2021 18:58
You can ask about it here.
REYNA_MAIN
17.12.2021 17:23
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Let the mid pt. of $AB$ be $M$ and $BC \cap MD = G$. Its easy to notice that $EF \parallel BC$.
Claim: Reflection of $D$ across $BC = M$.
$\angle DBC = \angle BAC = \angle ABC, \angle ADB = \angle AMB = 90^{\circ} \implies \angle DCB = \angle MCB, CDBM$ is cyclic $\implies \angle BDM = \angle BMD \implies \overline{BM} = \overline{BD} \implies \triangle BDG \cong \triangle BMG$ and we are done with our claim.
Now $\angle BEF = 180^{\circ} - \angle EBA, \angle DEB = \angle MEB = \angle MBE \implies \angle BED + \angle BEF = \angle MBE + 180^{\circ} - \angle EBA = 180^{\circ}$ and we are done .
Attachments:
lifeismathematics
09.11.2022 20:15
so mark up $\angle{CAB}=\angle{CBA}=A$ and $\angle{ACB}=C$
notice from alternate segment theorem $\angle{DBC}=A \implies \angle{DCB}=90^{\circ}-A$
also we get $C,D,B,F$ concylic
mark $FD\cap BC=E'$
notice we get $FE'\parallel AB$ so we get $FD\parallel AB$
Join $FE$ and notice $FEBA$ is a cyclic
in $\triangle{CEA} , \angle{CAE}=90^{\circ}-C$
so we get $\angle{EAB}=90^{\circ}-A\implies \angle{EFB}=90^{\circ}-A$
also since $BF \perp AC$ we have $\angle{E'FB}=90^{\circ}-A$
which forces $E'=E$ and hence we get $F,E,D$ to be collinear $\blacksquare$
dudade
27.06.2024 05:52
Notice, $CFBD$ is cyclic as $\angle CFB = \angle CDB = 90^{\circ}$. Let $O$ be the circumcenter, then \[ \angle BCO = \angle OBC = 90^{\circ} - \angle CBD = \angle BCD = \angle BFD. \]Now, $\angle BFE = 90^{\circ} - \angle CFE = \angle FCO = \angle BCO$. Therefore, $\angle BFE = \angle BFD$. Thus, $F$, $E$, and $D$ are collinear, as desired.
khanhnx
27.06.2024 06:27
Suppose that $EF$ intersects $\ell$ again at $D'$. We have $\angle{D'BC} = \angle{BAC} = \angle{EFC}$. Then $D', B, F, C$ lies on a circle. So $\angle{BD'C} = 90^{\circ}$ or $D' \equiv D$