Midpoints and foot of altitude are collinear

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Midpoints and foot of altitude are collinear

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Source: Centroamerican Olympiad 2016, problem 2

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Jutaro

#1 h Jun 19, 2016, 9:29 PM • 3 Y

Y by jbaca, faboyz0305, Adventure10

Let $ABC$ be an acute-angled triangle, $\Gamma$ its circumcircle and $M$ the midpoint of $BC$ . Let $N$ be a point in the arc $BC$ of $\Gamma$ not containing $A$ such that $\angle NAC= \angle BAM$ . Let $R$ be the midpoint of $AM$ , $S$ the midpoint of $AN$ and $T$ the foot of the altitude through $A$ . Prove that $R$ , $S$ and $T$ are collinear.

This post has been edited 2 times. Last edited by Jutaro, Jun 20, 2016, 8:18 PM
Reason: Typo in problem statement

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#2 h Jun 19, 2016, 9:52 PM • 3 Y

Y by lahmacun, Adventure10, Mango247

Hello, should $T$ be the foot of the altitude from $A$ ?

If so, it suffices to prove that $A',N,M$ are collinear, where $A'$ is the reflection of $A$ over $BC$ . We have $AMA'$ isoceles, so it suffices to prove $\angle AMB=\angle NMB$ .

There is a spiral similarity centered at $N$ taking $AB$ to $MC$ . To prove this, first note $\angle MCN=\angle BAN$ . We can then find $\angle MNC=\angle BNA$ by using trig/ratio lemma on $BNC$ (using $\frac{KB}{KC}=\frac{c^2}{b^2}$ where $K=AC\cap AN$ ). Then we have $\triangle NBM\sim NAC$ and clearly $\triangle NAC\sim \triangle BAM\implies \triangle NBM\sim BAM$ and $\angle AMB=\angle NMB$ as desired.

This post has been edited 1 time. Last edited by droid347, Jun 19, 2016, 10:51 PM
Reason: added solution

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#3 h Jun 19, 2016, 10:01 PM • 1 Y

Y by Adventure10

droid347 wrote:

Hello, should $T$ be the foot of the altitude from $A$ ?

Agreed, my current diagram is definitely not showing $R,S,T$ are collinear, $T$ being the foot of the altitude however makes sense.

This post has been edited 1 time. Last edited by Wave-Particle, Jun 19, 2016, 10:02 PM
Reason: oops

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#4 h Jun 19, 2016, 11:41 PM • 4 Y

Y by LoloVN, Adventure10, Mango247, Ksi2024

I consider $T$ the foot of the A-altitude:
Let $D$ be a point such that $DB$ and $DC$ are tangent to $\odot (ABC)$ $\Longrightarrow$ since $AN$ is A-symmedian we get $A$ , $N$ , $D$ are collinear. Let $E$ $=$ $AN$ $\cap$ $BC$ $\Longrightarrow$ $(A,N,E,D)=-1$ and $\measuredangle EMD$ $=$ $90^{\circ}$ combining these two results we get $\measuredangle AME$ $=$ $\measuredangle EMN$ , but since $TR$ $=$ $RM$ we get $\measuredangle RTM$ $=$ $\measuredangle RMT$ $\Longrightarrow$ $\measuredangle RTM$ $=$ $\measuredangle EMN$ $\Longrightarrow$ $TR\parallel MN...(1)$ , but since $S$ and $R$ are the midpoints of $AN$ and $AM$ we get $SR\parallel MN...(2)$ $\Longrightarrow$ by $(1)$ and $(2)$ we get $R,S,T$ are collinear.

This post has been edited 1 time. Last edited by FabrizioFelen, Jun 20, 2016, 3:25 PM

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PROF65

#5 h Jun 20, 2016, 2:18 AM • 1 Y

Y by Adventure10

Jutaro wrote:

Let $ABC$ be an acute-angled triangle, $\Gamma$ its circumcircle and $M$ the midpoint of $BC$ . Let $N$ be a point in the arc $BC$ of $\Gamma$ not containing $A$ such that $\angle NAC= \angle BAM$ . Let $R$ be the midpoint of $AM$ , $S$ the midpoint of $AN$ and $T$ the foot of the altitude through $A$ . Prove that $R$ , $S$ and $T$ are collinear.

if so then:
$AN$ is the symmedian,it s easy to see that $NM$ is the symmetric of the median wrt the $BC$ -bisector then also wrt $BC$ ,let $A'$ be the symmetric of $A$ wrt $BC$ thus $N,M,A'$ are collinear therefore $S,T,R$ are collinear.
R HAS

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ABCDE

#6 h Jun 20, 2016, 3:06 AM • 2 Y

Y by Adventure10, Mango247

Take a homothety with factor 2 at $A$ and the problem becomes $M$ , $N$ , and $A'$ , the reflection of $A$ over $BC$ , collinear. A $\sqrt{bc}$ inversion on $ABC$ swaps $M$ and $N$ and takes $A'$ to $O$ , the circumcenter of $ABC$ . An inversion about the circumcircle of $ABC$ fixes $A$ and $N$ and sends $M$ to the intersection of the tangents to the circumcircle at $B$ and $C$ , $D$ . Since $AND$ is a symmedian, $O$ lies on the circumcircle of $AMN$ , so $A'$ lies on $MN$ .

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#8 h Jun 20, 2016, 3:08 PM • 2 Y

Y by Adventure10, Mango247

$\angle RTM= \angle AMB= \angle MAC+ \angle ACB= \angle BAN+ \angle ANB= \angle BCN+ \angle MNC= \angle BMN$

$\Longrightarrow TR \parallel NM$

$\Longrightarrow TR$ passes through the midpoint of $AN$ which is $S$ .

Done!

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Jutaro

#9 h Jun 20, 2016, 8:07 PM • 1 Y

Y by Adventure10

I am sorry for the mistake. Indeed $T$ is meant to be the foot of the altitude through $A$ .

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#10 h Jun 21, 2016, 7:01 AM • 2 Y

Y by Adventure10, Mango247

Hello.

A different approach.

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Let $D$ be the intersection of $BC$ with the tangent to $\Gamma$ at $A$ and suppose that $Y\equiv AN\cap BC$ .

The line $AN$ is the $A-$ symmedian of $\triangle{ABC}$ .Hence,we know that the quadruple $D,Y/B,C$ is harmonic and that $AN$ is

the polar of $D$ with respect to $\Gamma$ (the above are well known but whoever needs more explanation may tell me).

Thus, $DS\perp AN$ which means that $ASTD$ is cyclic.It suffices to show that $AMND$ is also cyclic,because,then,

$\angle{MNA}=\angle{MDA}=\angle{MTS}\Rightarrow MN\parallel ST$ ,which is the desired result.Equivalently,it suffices to show that

$DY\cdot MY=AY\cdot NY\Leftrightarrow DY\cdot MY=BY\cdot CY\Leftrightarrow (DM-MY)MY=(BM-MY)(BM+MY)\Leftrightarrow$

$\Leftrightarrow MD\cdot MY=MB^2$ which is Newton's relation for the harmonic quadruple $D,Y/B,C$ .

This post has been edited 1 time. Last edited by gavrilos, Jun 21, 2016, 7:02 AM

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#11 h Jun 22, 2016, 2:27 AM • 3 Y

Y by mathisreal, Adventure10, Mango247

Consider $AM \cap \Gamma =D$
Since $RS||MN$ it's enough to prove that $TR||MN$

1) $R$ is the midpoint of $AM$
$\Rightarrow TR=AR=RM \Rightarrow \angle RTM = \angle RMT$

2) $\angle NAC = \angle BAM \Rightarrow \angle BAN = \angle MAC = \angle DAC$
then $BN=CD$

3) $\angle MBN = \angle MCD$ and $BM=MC \Rightarrow \triangle BMN \cong \triangle CMD$
( $SAS$ ) $\longrightarrow \angle BMN = \angle CMD = \angle AMB = \angle RMT$

4)Then, $\angle BMN = \angle RMT = \angle RTM \Rightarrow TR||MN$
$Q.E.D$

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#12 h Jun 22, 2016, 12:07 PM • 1 Y

Y by Adventure10

I found a tricky solution using spiral similarity twice and I thought this problem was too much for a 2nd problem in a omcc, but maybe the solution of Packito gives sense to the selection,...

This post has been edited 1 time. Last edited by hectorraul, Feb 1, 2017, 11:58 AM

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#13 h Jun 27, 2016, 7:44 PM • 1 Y

Y by Adventure10

¿Existe forma de resolverlo partiendo de que, por menelao en el triangulo AMX, con x el corte de AN con BC, y los puntos R,T y S, se tiene que para que sean colineales es porque AR/RM * MT/TX * XS/SA = 1, y como AR/RM=1, entonces MT/TX=SA/SX

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#14 h Jun 27, 2016, 7:55 PM • 2 Y

Y by jbaca, Adventure10

Using Menelaus? I think you outlined it. I think spiral similarity is more straight-forward.

This post has been edited 2 times. Last edited by abk2015, Jun 27, 2016, 7:57 PM

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#15 h Jun 28, 2016, 3:58 AM • 3 Y

Y by jbaca, Adventure10, Mango247

Let $A' \in \odot(ABC)$ such that $AA'\parallel BC$ and let $T'$ be the reflection of $A$ over $BC$ ; it's easy to see that $T', M, A'$ are collinear, so $-1=(B,C;M,P_{\infty,BC})\stackrel{A'}{=}(B,C;N,A)$ and a homothety at $A$ with ratio $\tfrac{1}{2}$ finishes the problem.

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jbaca

#16 h Sep 5, 2016, 6:50 AM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Hello. I'm the happy author of this problem. I hope everyone who participated in OMCC 2016 had overjoyed it. I used it as a lemma to solve a problem from Sharygin Geometry Olympiad, correspondence round.

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#17 h Sep 5, 2016, 7:54 AM • 2 Y

Y by Adventure10, Mango247

I am giving a solution;

assume AM intersects the circumcircle at D

here is RS || MN if R,S,T is collinear, it should be RT || MN

in triangle RTM , R is the midpoint of hypontenous AM so ,easily AR = RM = RT;

<RTM = <RMT (1)

<RTM = <CMN = 180 - <BMN;

<RMT = 180 - <AMB = 180 - <CMD;

from (1) ,we need to prove that, <BMN = <CMD;

<NAC = <BAM;

or, <CAD = <BAN;

from sine law, BN = CD;

also, <CAN = <BAM;

or, <CAN = <BAD;

or, <CBN = <BCD;

or, <MBN = <MCD;

in triangle BMN and CMD,

BM = CM;

BN = CD;

and <MBN = <MCD;

so, triangle BMN and triangle CMD is simmilar,

now, <BMN = <CMD

this proofs the concurrency of R,S,T

btw my first solution of higher olympiad level geometry as centroamerican

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#18 h Sep 5, 2016, 8:06 AM • 2 Y

Y by Adventure10, Mango247

brothers can you give me any idea about homothaty and reflection in mathmetics

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#19 h Mar 1, 2019, 4:31 PM • 2 Y

Y by Pluto1708, Adventure10

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.653987447520949, xmax = 21.54578172460581, ymin = -17.457492228471796, ymax = 4.6489279762757745; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-3.7012747299195645,3.4292634132552204)--(-6.08,-3.99)--(5.04,-4.15)--cycle, linewidth(4) + rvwvcq); /* draw figures */ draw((-3.7012747299195645,3.4292634132552204)--(-6.08,-3.99), linewidth(4) + rvwvcq); draw((-6.08,-3.99)--(5.04,-4.15), linewidth(4) + rvwvcq); draw((5.04,-4.15)--(-3.7012747299195645,3.4292634132552204), linewidth(4) + rvwvcq); draw(circle((-0.48579318418222933,-1.692626300664937), 6.047567760061765), linewidth(2.8)); draw((-3.7012747299195645,3.4292634132552204)--(0.9623058187584692,-7.564260237672644), linewidth(2.8)); draw((-2.1022256244814166,-7.520166260072069)--(-3.7012747299195645,3.4292634132552204), linewidth(2) + wrwrwr); draw((-3.7012747299195645,3.429263413255214)--(xmin, 69.5000000000002*xmin + 260.6678571426657), linewidth(2.8)); /* ray */ draw((-6.08,-3.99)--(-0.7070972157740317,-17.073256496295194), linewidth(2.8)); draw((-0.7070972157740317,-17.073256496295194)--(5.04,-4.15), linewidth(2.8)); draw((-0.7070972157740317,-17.073256496295194)--(-3.7012747299195645,3.4292634132552204), linewidth(3.2)); draw((-0.52,-4.07)--(-0.7070972157740317,-17.073256496295194), linewidth(2.8) + dtsfsf); draw((-0.52,-4.07)--(xmin, 2.1805779192856143*xmin-2.93609948197148), linewidth(2.8) + dtsfsf); /* ray */ /* dots and labels */ dot((-3.7012747299195645,3.4292634132552204),linewidth(6pt) + dotstyle); label("$A$", (-4.463565081807408,3.886637624387928), NE * labelscalefactor); dot((-6.08,-3.99),linewidth(6pt) + dotstyle); label("$B$", (-7.055352278226076,-4.4070814041518505), NE * labelscalefactor); dot((5.04,-4.15),linewidth(6pt) + dotstyle); label("$C$", (5.38522626458353,-4.6815059308314755), NE * labelscalefactor); dot((-0.52,-4.07),dotstyle); label("$M$", (-0.19473911123548462,-3.705774280415031), NE * labelscalefactor); dot((0.9623058187584692,-7.564260237672644),dotstyle); label("$M'$", (0.537059626576845,-8.49295769027071), NE * labelscalefactor); dot((-2.110637364959782,-0.3203682933723899),dotstyle); label("$R$", (-1.7498114290866853,-0.016288977277850253), NE * labelscalefactor); dot((-3.7012747299195623,3.429263413255221),dotstyle); dot((-2.1022256244814166,-7.520166260072069),dotstyle); label("$N$", (-3.1219340624848035,-8.035583479138003), NE * labelscalefactor); dot((-2.9017501772004906,-2.0454514234084247),dotstyle); label("$S$", (-2.6035766232010698,-1.784802593657656), NE * labelscalefactor); dot((-3.8084969876150505,-4.022683496581078),dotstyle); label("$T$", (-4.616023152184977,-4.803472387133531), NE * labelscalefactor); dot((-3.915719245310536,-11.474630406417367),dotstyle); label("$D$", (-4.920939292940114,-11.969001694879294), NE * labelscalefactor); dot((-0.7070972157740317,-17.073256496295194),dotstyle); label("$K_A$", (-0.5911300942171632,-16.756185104734975), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $MN \cap AT=D$ , and let the tangents at $B$ and $C$ intersect at $K_A$ ,
$-1=(A,N; AN \cap BC,K_A) \overset{M}{=} (A,D;T, \infty_{MK_A}) \Longrightarrow AT=TD$ and then finish off using mid-point theorem

This post has been edited 1 time. Last edited by AlastorMoody, Mar 1, 2019, 4:31 PM

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Synthetic_Potato

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Synthetic_Potato

#20 h Mar 2, 2019, 10:09 AM • 2 Y

Y by AlastorMoody, Adventure10

Let $A'$ be the reflection of $A$ over $BC$ . It suffices to show that $M,N,A'$ are collinear. We know that the reflection of the $A-$ HM point over $BC$ is $N$ . Thus it follows that if $N'$ is the reflection of $N$ over $BC$ then $N'\in \odot(A'BC)$ . Also, $N'\in AM$ from definition. Reflecting $A,N',M$ over $BC$ yields that $A',M,N$ are collinear. $\blacksquare$ .

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#21 h Mar 2, 2019, 10:52 AM • 2 Y

Y by Adventure10, Mango247

Here is my solution for this problem
Solution
Let $P$ $\in$ ( $O$ ) such as $AT$ $\parallel$ $BC$ ; $A'$ be reflection of $A$ through $BC$
We have: 1 = $\dfrac{MB}{MC}$ = $\dfrac{AC}{AB}$ . $\dfrac{NB}{NC}$ = $\dfrac{PB}{PC}$ . $\dfrac{NB}{NC}$ or $N$ , $M$ , $P$ are collinear
But: $\dfrac{A'T}{A'A}$ = $\dfrac{TM}{AP}$ = $\dfrac{1}{2}$ so: $A'$ , $M$ , $P$ are collinear
Then: $P$ , $M$ , $N$ , $A'$ are collinear or $R$ , $S$ , $T$ are collinear

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#22 h Mar 2, 2019, 7:02 PM • 2 Y

Y by Adventure10, Mango247

I've got synthetic one. I used notations as in diagram in post #10
$AB=c\wedge BC=a\wedge CA=b$ We want to prove $MN||ST$ which is $\frac{MY}{NY}=\frac{TY}{SY}$
Line $AN$ is $A-$ symedian so $BY=\frac{ac^2}{b^2+c^2}\wedge CY=\frac{ab^2}{b^2+c^2}$
$MY=|\frac{a}{2}-\frac{ab^2}{b^2+c^2}|=\frac{a|b^2-c^2|}{2(b^2+c^2)}$
In $BT$ exceptionally the length is signed, so we can work even if $ABC$ isn't acute
$BT=c\cos\angle ABC=\frac{a^2+c^2-b^2}{2a}$
$TY=|BY-BT|=|\frac{ac^2}{b^2+c^2}-\frac{a^2+c^2-b^2}{2a}|=\frac{|b^2-c^2||a^2-b^2-c^2|}{2a(b^2+c^2)}$
Concyclicity yields $AY\cdot NY=BY\cdot CY=\frac{a^2b^2c^2}{(b^2+c^2)^2}$
$SY=|NS-NY|=\frac{|AY-NY|}{2}$ which is true
By Stewart's theorem $AY^2=\frac{BY\cdot AC^2+CY\cdot AB^2}{BC}-BY\cdot CY=\frac{2b^2c^2}{b^2+c^2}-\frac{a^2b^2c^2}{(b^2+c^2)^2}$
$\frac{MY}{NY}=\frac{TY}{SY}\iff \frac{\frac{a|b^2-c^2|}{2(b^2+c^2)}}{NY}=\frac{\frac{|b^2-c^2||a^2-b^2-c^2|}{2a(b^2+c^2)}}{\frac{|AY-NY|}{2}}\iff \frac{a^2}{ AY\cdot NY}=\frac{2|a^2-b^2-c^2|}{|AY^2-NY\cdot AY|}\iff$
$\frac{a^2}{\frac{a^2b^2c^2}{(b^2+c^2)^2}}=\frac{2|a^2-b^2-c^2|}{|\frac{2b^2c^2}{b^2+c^2}-\frac{a^2b^2c^2}{(b^2+c^2)^2}-\frac{a^2b^2c^2}{(b^2+c^2)^2}|}\iff \frac{1}{b^2c^2}=\frac{2|a^2-b^2-c^2|}{2b^2c^2|b^2+c^2-a^2|}$

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#24 h Apr 1, 2021, 5:58 PM

Y by

It is clear that $AN$ is the $A$ -symmedian of $\triangle ABC$ .
Let $A'=AT\cap MN$ . Also let $R$ and $K$ be the intersections of the $\angle A$ bisector with $(ABC)$ and $BC$ respectively.
By Russia 2009 $NMKR$ is cyclic.
So by
$\angle MA'T=90-\angle NMK=90-\angle NCA=90-\angle C-\angle CAM=\angle AMC-90=\angle MAT$ .
we get that $T$ is the midpoint of $AA'$ and by homothety we are done.

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#25 h Dec 24, 2022, 3:16 AM

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First, $\angle MAC + \angle NAM = \angle NAC = \angle BAM = \angle BAN + \angle NAM$ $\Rightarrow$ $\angle BAN = \angle MAC$ .
In the $\triangle AMN$ we have that $\frac{AS}{AN} = \frac{AR}{AM} = \frac{1}{2}$ , so by Thales's theorem we get that $SR \parallel MN$ .
As $\angle BAM = \angle NAC$ and $\angle ANC = \angle ABC$ by case $AA$ , $\triangle BAM$ $\sim$ $NAC$ . Then,
$\frac{MA}{AC} = \frac{BM}{NC} = \frac{MC}{CN}$ .
Notice that $\angle BAN = \angle BCN = \angle MAC$ so by case $LAL$ we get that $\triangle MAC \sim \triangle MCN$ . Hence, $\angle AMC = \angle NMC = 180^{\circ} - \angle BMA = 180^{\circ} - \angle BMN \Rightarrow \angle BMA = \angle BMN$
We see that $R$ Is circumcenter of $\triangle ATM$ $\Rightarrow$ $RA = RT = RM \Rightarrow \angle TMR = \angle MTR = \angle BMA = \angle BMN$ . Hence, $\angle ART = 2 \angle AMT = 2 \angle BMA = \angle BMA + \angle BMN = \angle AMN$
It follows that $TR \parallel MN$ and since $SR \parallel MN$ , then $R$ , $S$ and $T$ lies on the same line parallel to $MN$ $\blacksquare$

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#26 h Dec 24, 2022, 3:40 AM • 1 Y

Y by Mango247

Let $P_A$ be the $A$ Humpty point of triangle $ABC$ , $A'$ is the reflection of $A$ over line $BC$ . We have $P_A$ is the reflection of $N$ over line $BC$ . Thus, $M,N,A'$ are collinear. Then, $TS//A'N, TR//A'M \Rightarrow R,S,T$ are collinear.

This post has been edited 1 time. Last edited by faboyz0305, Dec 24, 2022, 3:42 AM

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