Symmedian line

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April

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April

#1 h May 10, 2009, 3:49 PM • 8 Y

Y by Amir Hossein, dave_mathsguru, 799786, itslumi, Adventure10, Mango247, fuzzball2023, and 1 other user

Let be given a triangle $ABC$ and its internal angle bisector $BD$ $(D\in BC)$ . The line $BD$ intersects the circumcircle $\Omega$ of triangle $ABC$ at $B$ and $E$ . Circle $\omega$ with diameter $DE$ cuts $\Omega$ again at $F$ . Prove that $BF$ is the symmedian line of triangle $ABC$ .

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mathVNpro

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mathVNpro

#2 h May 10, 2009, 5:14 PM • 6 Y

Y by Olemissmath, Psyduck909, Adventure10, Mango247, fuzzball2023, ehuseyinyigit

Let $M_b$ be midpoint of $AC$ .
We have $\angle FAM_b = \angle FBC$ , $\angle AM_bF\equiv \angle DM_bF = \angle DEF = \angle BEF = \angle BCF$
$\Rightarrow \triangle FAM_b\sim \triangle FBC$ , $(i)$ . Thus, through the spiral similarity with center $F$ , denote this $f$ then:
$f: \triangle FAM_b\mapsto \triangle FBC$
$\Rightarrow f: A\mapsto B$ , $M_b\mapsto C$
$\Rightarrow f: A\mapsto M_b$ , $B\mapsto C$
$\Rightarrow f: \triangle FAB\mapsto \triangle FM_bC$ , $(1)$
But it is easy to notice that: $\triangle BAF\sim \triangle BM_bC$ (a.a), $(2)$
$(1),(2)\Rightarrow \triangle CM_bB\sim \triangle FM_bC$
$\Rightarrow \frac {CB}{CF} = \frac {CM_b}{FM_b} = \frac {M_bB}{M_bC}$ , $(*)$ .
With the same argument, we get $\triangle BAM_b\sim \triangle BFC$
$\Rightarrow \triangle BAM_b\sim \triangle AFM_b$ (thanks to $(i)$ ).
$\Rightarrow \frac {BA}{AF} = \frac {M_bB}{M_bA}$ , $(**)$
But $M_bA = M_bC \Rightarrow \frac {BA}{BC} = \frac {FA}{FC}$ (Combine $(*)\&(**)$ )
$\Longrightarrow ABCF$ is a harmonic quadrilateral. Therefore, $BF$ passes through the Pole of $AC$ wrt $(O)$ , where $O$ is the circumcenter of $\triangle ABC$ . Then we will get the result of the problem.
Our proof is completed then

This post has been edited 1 time. Last edited by mathVNpro, May 11, 2009, 4:56 AM

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silouan

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silouan

#3 h May 10, 2009, 6:41 PM • 6 Y

Y by ultralako, Adventure10, Mango247, fuzzball2023, and 2 other users

Let $M$ the midpoint of $AC$ , $O$ the circumcenter of $\Omega$ and let the tangent to $\Omega$ at $B$ meets $AC$ at $R$ .
Then $BOMR$ is cyclic (let $Q$ its circumcenter) . Then $R,Q,O$ are collinear . So if the circumcircle of $BOMR$ meets $\Omega$ at $F'$ , then $BF'\bot RO$ . So by polars $RF'$ is tangent to $\Omega$ . By well known theorem we have that $BF'$ is the $B$ symmedian. Finally if $BE$ meets the circumcircle of $BOMR$ at $S$ , then $SM=SF$ . But also because $RF'=RB$ we have that $D$ is the incenter of $F'BM$ so $SD=SM=SF$ and because $DM\bot ME$ we also have $SD=SM=SF=SE$ . This shows that $MDF'E$ is cyclic and so $F\equiv F'$ and we are done .

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yetti

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yetti

#4 h May 10, 2009, 6:54 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

External bisector of $\angle B$ cuts $CA$ at $Y.$ $I, I_b$ are incenter and B-excenter of $\triangle ABC.$ $(E)$ is a circle with center $E$ and radius $EA = EC = EI = EI_b.$ The cross ratio $(B, D, I, I_b)$ is harmonic and $E$ the midpoint of $II_b$ $\Longrightarrow$ $\overline{BD} \cdot \overline{BE} = \overline{BI}\cdot \overline{BI_b}$ $\Longrightarrow$ $BY \perp BDE$ is radical axis of $\omega, (E).$ Pairwise radical axes $EF, CA, BY$ of $\Omega, \omega, (E)$ meet at their radical center $Y.$ Since $DF \perp EFY$ and $BY \perp BDE,$ $BDFY$ is cyclic. Its circumcircle is the B-Apollonius circle of $\triangle ABC$ $\Longrightarrow$ $BF$ is the B-symmedian.

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Virgil Nicula

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Virgil Nicula

#5 h May 11, 2009, 2:26 AM • 6 Y

Y by Amir Hossein, Adventure10, Mango247, fuzzball2023, and 2 other users

PP. Let be given a triangle $ABC$ and its internal angle bisector $BD$ $(D\in BC)$ . The line $BD$ intersects the circumcircle $\Omega$ of
$\triangle ABC$ at $B$ and $E$ . Circle $\omega$ with diameter $DE$ cuts $\Omega$ again at $F$ . Prove that $BF$ is the symmedian line of triangle $ABC$ .

Lemma. Let $ABC$ be a triangle with the circumcircle $w$ . Denote the following points : the middlepoint $M$ of the side $[BC]$ ;

the point $D\in (BC)$ for which $\angle DAB\equiv\angle DAC$ ; the point $E$ for which $\{A,E\} = AD\cap w$ ; the $A$ -symmedian $AL$ ,

where $L\in (BC)$ ; the point $S$ for which $\{A,S\} = AL\cap w$ . Then $SD\perp SE$ , i.e. the quadrilateral $MDSE$ is cyclically.

Proof. Denote $XX$ - the tangent in $X\in w$ to $w$ . Denote $T\in BB\cap CC$ . Observe that $\overline {MET}\perp BC$ and $\angle ASB\equiv\angle ACB$ .

From the first well-known property, $T\in \overline {ALS}$ and the line $\overline{ALST}$ is the $A$ -symmedian in $\triangle ABC$ $\Longrightarrow$ $\left\|\begin{array}{c} \angle BAS\equiv\angle MAC \\ \ \angle SAE\equiv\angle DAM\end{array}\right\|$ .

From the second well-known property, the division $\{A,S;L,T\}$ is harmonically. Since $ML\perp MT$ results $\angle DMA\equiv\angle DMS$ .

Show easily that $\triangle BAS\sim\triangle MAC$ (a.a.) . Thus, $\frac {AB}{AS} = \frac {AM}{AC}$ , i.e. $AS\cdot AM = AB\cdot AC$ .

From the third well-known property $AD\cdot AE = AB\cdot AC$ , obtain $\boxed {AS\cdot AM = AD\cdot AE = AB\cdot AC}$ , i.e. $\frac {AS}{AE} = \frac {AD}{AM}$ .

Since $\angle SAE\equiv\angle DAM$ obtain $\triangle SAE\sim\triangle DAM$ and in consequence, $\angle DMA\equiv\angle SEA$ . Since $\angle DMA\equiv\angle DMS$

and $\angle SEA\equiv\angle SED$ obtain $\angle DMS\equiv\angle SED$ , i.e. the quadrilateral $MDSE$ is cyclically. In conclusion, $SD\perp SE$ .

Remark. Here are another interesting metrical relations : $\left\|\begin{array}{ccc} ABS\sim CMS & \implies & \boxed {SB\cdot SC = SA\cdot SM} \\ \\ AMC\sim CMS & \implies & \boxed {MA\cdot MS = MB^2}\end{array}\right\|$ .

The proposed problem is an immediate consequence of the above lemma. See PP13 from here

This post has been edited 4 times. Last edited by Virgil Nicula, Aug 11, 2014, 1:54 AM

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jayme

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jayme

#6 h May 11, 2009, 5:22 AM • 3 Y

Y by AlastorMoody, Adventure10, fuzzball2023

Dear Mathlinkers,
1. Let A' be the midpoint of BC, X, Y the second meetpoint of AA', EA' with the circumcircle of ABC and Te the tangent to this circumcircle at E.
2. Te // DA'
3. The circle with diameter ED passes through A'
4. By a converse of Reim's theorem applied to this circle and the circumcircle, FD goes through Y;
5. By a converse of Pascal's theorem applied to EAXFYTe, FX //DA' (BC)
and we are done...
Sincerely
Jean-Louis

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Agr_94_Math

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Agr_94_Math

#7 h May 11, 2009, 6:19 AM • 3 Y

Y by Adventure10, Mango247, ehuseyinyigit

Very nice application of Reim's Theorem jayme.Though a very obvious result, it has great application in many synthetic geometry results.
Thanks.

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mathVNpro

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mathVNpro

#8 h May 11, 2009, 10:26 AM • 2 Y

Y by Adventure10, Mango247

jayme wrote:

Dear Mathlinkers,
1. Let A' be the midpoint of BC, X, Y the second meetpoint of AA', EA' with the circumcircle of ABC and Te the tangent to this circumcircle at E.
2. Te // DA'
3. The circle with diameter ED passes through A'
4. By a converse of Reim's theorem applied to this circle and the circumcircle, FD goes through Y;
5. By a converse of Pascal's theorem applied to EAXFYTe, FX //DA' (BC)
and we are done...
Sincerely
Jean-Louis

Do you have any document about Reim's theorem, Jayme? I am so excited about this.

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jayme

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jayme

#9 h May 11, 2009, 11:29 AM • 3 Y

Y by Adventure10, Mango247, fuzzball2023

See for example my website
http://perso.orange.fr/jl.ayme then: A propos
You will see all the possibility.
Sincerely
Jean-Louis

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ma 29

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ma 29

#10 h May 11, 2009, 12:02 PM • 16 Y

Y by MexicOMM, Wizard_32, Durjoy1729, HolyMath, Siddharth03, Greenleaf5002, myh2910, snakeaid, centslordm, Jalil_Huseynov, Adventure10, Mango247, GustavoPudim, fuzzball2023, and 2 other users

Dear friends.
I have a short solution for the nice problem,here

.

//cdn.artofproblemsolving.com/images/4a3755e6df266a6313f50f6a8a6c7c8845715061.jpg

Denote by $A'$ the midpoint of segment $AC$ .

$EO$ meets $(O)$ again at $P$

By ${ \angle {DFE} = 90^0}$ ,we have $F,D,P$ are collinear.

From $BPEF$ is cyclic and $BDA'P$ is cylic ,we have :

$\angle {EBF} = \angle {EPF} = \angle{DBA'}$ .

Therefor, $BF$ is the symmedian line of the triangle $BAC$ .

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armpist

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armpist

#11 h May 11, 2009, 1:38 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

All are very nice solutions indeed, I must admit.

T.Y.
M.T.

This post has been edited 1 time. Last edited by armpist, May 11, 2009, 8:29 PM

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Virgil Nicula

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Virgil Nicula

#12 h May 11, 2009, 4:21 PM • 2 Y

Y by Adventure10, Mango247

After the Ma29's shortest proof, this problem seems very simple ...
In the another proofs appeared many nice properties of this configuration.
Thanks to all for their interest and proofs.

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ma 29

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ma 29

#13 h May 13, 2009, 8:58 AM • 2 Y

Y by Adventure10, Mango247

Hello my friends.
I think all proof above are very nice.

Here is another solution to this problem:
//cdn.artofproblemsolving.com/images/c51085c9ea4036fa17fa17def72321254472bf4e.jpg

Denote $V = EF \cap AC$ .

We have : $\bar {VD} .\bar{VA'} = \bar{VF}.\bar{VE} = \bar{VA}.\bar{VC}$ (1)

But $A'$ is the midpoint of segment $AC$ .(2)

From (1) ,(2) and Maclaurin's theorem , we obtain the division ${VDAC}$ is harmonically,so $V,B,P$ are collinear.

Denote $Q = BC\cap {PA}$ and denote by $H$ the intersection of two tangent to $(O)$ at $A,C$ ,respectively.

We have $Q,D,H$ lie on the polar of $V$ wrt $(O)$ ,i.e $Q,D,H$ are collinear.

By Pascal's theorem applied to six point $A,A,P,C,B,F$ ,we obtain $B,F,H$ are collinear.

Therefore , $BF$ is the symedian line of the triangle $ABC$ .

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bravado

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bravado

#14 h May 13, 2009, 10:30 PM • 2 Y

Y by Adventure10, Mango247

It can also be viewed as an easy application of : http://www.mathlinks.ro/viewtopic.php?p=1082899#1082899

Indeed, from the above problem it follows that the line $FM$ (where $M$ is the midpoint of $AC$ ) meets the circumcircle of $\triangle{ABC}$ at a point $T$ such that $BT||AC$ . By symmetry,
$\angle{MBC} = \angle{MTA} = \angle{ABF}.$

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Virgil Nicula

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Virgil Nicula

#15 h May 14, 2009, 8:32 PM • 2 Y

Y by Adventure10, Mango247

We can use my found property at the proposed problem from here.

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shendrew7

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shendrew7

#80 h Dec 27, 2023, 10:31 PM

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Let $M$ be the midpoint of $BC$ and $N$ be the top point of $(ABC)$ . Then

$\angle EFD = \angle EFN = 90$ , so $NDF$ collinear.
$\angle NAE = \angle DMN$ , so $NADM$ is cyclic.

Thus $AF$ is a symmedian, as
$\measuredangle MAD = \measuredangle MND = \measuredangle DAF. \quad \blacksquare$

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kamatadu

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kamatadu

#81 h Dec 30, 2023, 6:21 PM

Y by

We solved the $A$ -centered version of this problem.

Perform a $\sqrt{bc}$ -inversion. Then note that $D\leftrightarrow E$ which means that $\odot(DE)$ is fixed under this inversion. Now $BC \leftrightarrow \odot(ABC)$ and thus, $F\leftrightarrow \odot(DE)\cap BC$ . But note that $\odot(DE)\cap BC$ is simply the midpoint of $BC$ . Thus $AF$ is isogonal to the $A$ -median and so, $AF$ is the $A$ -symmedian. :yoda:

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AshAuktober

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AshAuktober

#82 h Mar 1, 2024, 3:27 PM

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Perform a force-overlaid inversion (complete with the reflection about the angle bisector) about $B$ . Let the resultant image of any object $X$ be $X'$ . Note that $F', E'$ must lie on line $A'C'$ , and also that $90^\circ = \angle DFE = \angle BFE - \angle BFD - \angle BE'F' - \angle BD'F = \angle E'F'D'$ ; and thus $D'F' \perp A'C'$ ; however, as $D'$ is the midpoint of arc $A'C'$ , $F'$ is nothing but the midpoint of $A'C'$ . Thus $BF$ is the reflection of the median about the angle bisector, i.e. the symmedian. $\square$

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Aiden-1089

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Aiden-1089

#83 h Mar 3, 2024, 3:56 AM

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Let $E'$ be the antipode of $E$ in $(ABC)$ . Since $\angle EBE' = 90 ^{\circ}$ , we have $(A,C;D,BE' \cap AC)=-1$ .
Also, $F$ lies on $DE'$ , so $(A,C;F,B) \stackrel{E'}{=} (A,C;D,BE' \cap AC) = -1$ , which implies that $BF$ is a symmedian of $\Delta ABC$ .

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MagicalToaster53

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MagicalToaster53

#84 h Jul 7, 2024, 8:51 PM

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Let $\omega$ denote the circle with diameter $\overline{DE}$ . Suppose $\omega$ meets $\overline{BC}$ at $M$ other than $D$ . Then observe that $M$ is the midpoint of $\overline{BC}$ as $E$ bisects minor arc $\widehat{BC}$ and $\measuredangle DME = 90^{\circ}$ .

Now consider the inversion at $A$ with radius $\sqrt{bc}$ followed by a reflection about $AD$ , the angle bisector of $\angle BAC$ . Then $D$ maps to $E$ under this inversion, and it evidently follows that the image of $M$ is $F$ . Hence $AM$ and $AF$ are isogonal, as desired. $\blacksquare$

Remark

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Ywgh1

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Ywgh1

#85 h Aug 23, 2024, 2:00 PM

Y by

Russia 2009

Consider a $\sqrt{bc}$ inversion, we have that $D,E$ will be swapped, hence circle with diameter $DE$ fixed, $F$ will be sent to foot to $BC$ which is the mid of $BC$ hence $AF$ is Symedian

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Mathandski

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Mathandski

#86 h Sep 26, 2024, 4:52 PM

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$\text{[math]}$

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Anshu_Singh_Anahu

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Anshu_Singh_Anahu

#87 h Dec 11, 2024, 8:37 AM

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In the Following diagram

Define A function F(Z) = ±(BZ/CZ) , Now clearly BE = EC so F(E) = -1 aswell H is midpt of BC so F(H)= 1 also by angel bisector theorem F(D)= F(A) , by two circle intersecting at BC lemma F(H) × F(D) = F(E) × F(G) so ,F(G) = F(D) = F(A) , also F(G) × F(A) = F(M) , hence F(M) = (F(A))^2 and we are dn with AM as the symmedian cute~

I don't have cared about the direct length but no problem ig

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Anshu_Singh_Anahu

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Anshu_Singh_Anahu

#88 h Dec 11, 2024, 8:41 AM

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Above sol is too short ig it may be wrong check someone please :blush:

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lelouchvigeo

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lelouchvigeo

#89 h Dec 11, 2024, 10:21 AM

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Sol

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Ilikeminecraft

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Ilikeminecraft

#90 h Feb 21, 2025, 3:30 AM

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Note that the midpoint of $BC, M,$ passes through $(DE).$ Take $\sqrt{bc}$ inversion, and we get that $\omega$ maps to itself. Thus, $K \leftrightarrow M,$ so $AK$ and $AM$ are isogonal.

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TestX01

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TestX01

#91 h Feb 21, 2025, 10:51 AM

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ma 29 wrote:

http://i524.photobucket.com/albums/cc322/khanhto-photo/untitled-108.jpg

borrowing points here. Redefine $F$ as on symmedian. RTP $F,D,P$ collinear. DDIT on $A'FDE$ means that $EF\cap AC$ and $DF\cap EA'$ are isogonal from existing isogonal pairs $BD,BE$ and $BF, BA'$ . RTP now $EF, BP,AC$ concur clearly. This is as $-1=(B,F;A,C)$ is projected to $(D,EF\cap AC,A,C)$ . But well-known by apollonian that $BP\cap AC$ is harmonic conj of $D$ , so coincide with $EF\cap AC$ , yay.

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Avron

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Avron

#92 h Apr 7, 2025, 2:58 PM

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Let M, N be midpoints of BC and arc BAC, then $N,M,E$ are collinear and $ADMN$ is cyclic. Note that $\angle EFD=90=\angle EFN$ so $F,D,N$ are also collinear and we get:
$\angle FAD=\angle FAE = \angle FNE = \angle DNM=\angle DAM$ and we're done since $AD$ is the angle bisector

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bjump

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bjump

#93 h Apr 29, 2025, 12:43 AM

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Today is the day I finally understand how force overlaid inversion can actually be useful.
Perform a Force-Overlaid inversion with respect to $A$ . Note that $D \leftrightarrow E$ , so the diameter of $(DE)$ is fixed, therefore $F$ is sent to the midpoint of $BC$ finishing.

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BS2012

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BS2012

#94 h Apr 29, 2025, 12:58 AM

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This problem is 2012 AIME II P15 $-$ 2024 AIME I P10 explanation

Also slightly off topic but force overlay inversions are also useful to prove that the config in 2024 aime p10 is in fact a symmedian

This post has been edited 1 time. Last edited by BS2012, Apr 29, 2025, 1:05 AM

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