Difference between revisions of "Symmedians, Lemoine point"

Revision as of 14:46, 4 August 2024

The reflecting of the median over the corresponding angle bisector is the symmedian. The angle formed by the symmedian and the angle bisector has the same measure as the angle between the median and the angle bisector, but it is on the other side of the angle bisector. The symmedian $AS_A$ is isogonally conjugate to the median $AM_A.$

There are three symmedians. They are meet at a triangle center called the Lemoine point.

1 Proportions
2 Radical axis of circumcircle and Apollonius circle
3 Symmedian and tangents
4 Lemoine point properties
5 Parallel lines
6 Radical axis
7 Construction of symmedian’s point
8 Common Lemoine point
9 Lemoine point extreme properties
10 Lemoine point and perpendicularity
11 Lemoine point line
12 Antiparallel lines and segments
13 Three intersecting antiparallel segments
14 Three intersecting parallel to sides segments
15 Tucker circle
16 Tucker circle 2

Proportions

Let $\triangle ABC$ be given.

Let $AM$ be the median, $\Omega = \odot ABC, E \in BC, D = AE \cap \Omega \ne A.$

Prove that iff $AE$ is the symmedian than $\frac {BD}{CD} = \frac{AB}{AC}, \frac {BE}{CE} = \left (\frac{AB}{AC} \right )^2.$

Proof

1. Let $AE$ be the symmedian. So $\angle BAD = \angle CAM.$ $\angle BDA = \angle ACB = \angle ACM \implies \triangle ABD \sim \triangle AMC \implies$ $\frac {AM}{MC}= \frac {AB}{BD}.$ Similarly $\triangle ABM \sim \triangle ADC \implies \frac {AM}{MB}= \frac {AC}{CD}.$ $BM = MC \implies \frac {BD}{CD} = \frac{AB}{AC}.$

By applying the Law of Sines we get $\frac{AB}{\sin \angle AEB} = \frac{BE}{\sin \angle BAD}, \frac{CD}{\sin \angle CED} = \frac{CE}{\sin \angle CDE},$ $\frac{AC}{\sin \angle ADC} = \frac{BD}{\sin \angle BAD} \implies \frac {BE}{CE} = \frac{AB}{CD} \cdot \frac {BD}{AC} = \frac{AB^2}{AC^2}.$ Similarly, $\frac {AE}{ED} = \frac{AB^2}{BD^2}.$

2. $\frac {BD}{CD} = \frac{AB}{AC}.$

As point $D$ moves along the fixed arc $BC$ from $B$ to $C$ , the function $F(D) = \frac {BD}{CD}$ monotonically increases from zero to infinity. This means that there is exactly one point at which the condition is satisfied. In this case, point $D$ lies on the symmedian.

Similarly for point $E.$

Corollary

Let $AE$ be the $A-$ symmedian of $\triangle ABC.$

Then $BE$ is the $B-$ symmedian of $\triangle ABD, CE$ is the $C-$ symmedian of $\triangle ACD, DE$ is the $D-$ symmedian of $\triangle BCD.$

vladimir.shelomovskii@gmail.com, vvsss

Radical axis of circumcircle and Apollonius circle

The bisectors of the external and internal angles at vertex $A$ of $\triangle ABC$ intersect line $BC$ at points $D$ and $E.$ The circle $\omega = \odot ADE$ intersects the circumcircle of $\triangle ABC$ at points $A$ and $Y.$ Prove that line $AY$ contains the $A-$ symmedian of $\triangle ABC.$

Proof

The circle $\omega$ is the Apollonius circle for points $B$ and $C \implies$ $\frac {BE}{CE} = \frac {AB}{AC} = \frac {BY}{CY} \implies$ $AY$ is the $A-$ symmedian of $\triangle ABC.$

vladimir.shelomovskii@gmail.com, vvsss

Symmedian and tangents

Let $\triangle ABC$ and it’s circumcircle $\Omega$ be given.

Tangents to $\Omega$ at points $B$ and $C$ intersect at point $F.$

Prove that $AF$ is $A-$ symmedian of $\triangle ABC.$

Proof

Denote $D = AF \cap \Omega \ne A.$ WLOG, $\angle BAC < 180^\circ.$ $\triangle FDB \sim \triangle FBA \implies \frac {BD}{AB} = \frac{DF}{BF}.$ $\triangle FDC \sim \triangle FCA \implies \frac {CD}{AC} = \frac{DF}{CF}.$ $BF = CF \implies \frac {BD}{CD} = \frac{AB}{AC} \implies AD$ is $A-$ symmedian of $\triangle ABC.$

Corollary

Let $\triangle ABC$ and it’s circumcircle $\Omega$ be given.

Let tangent to $\Omega$ at points $A$ intersect line $BC$ at point $F.$

Let $FD$ be the tangent to $\Omega$ different from $FA.$

Then $AD$ is $A-$ symmedian of $\triangle ABC.$

vladimir.shelomovskii@gmail.com, vvsss

Lemoine point properties

Let $\triangle ABC$ be given. Let $L$ be the Lemoine point of $\triangle ABC.$

$LD \perp BC, D \in BC, LE \perp AC, E \in AC, LF \perp AB, F \in AB.$

Prove that $\frac{LD}{BC} = \frac{LE}{AC} = \frac{LF}{AB}, L$ is the centroid of $\triangle DEF.$

Proof

Let $G$ be the centroid of $\triangle ABC, GD' \perp BC, D' \in BC,$ $LE' \perp AC, E' \in AC, LF' \perp AB, F' \in AB.$

The double area of $\triangle AGC$ is $2[AGC] = GE' \cdot AC = 2[BGC] = GD' \cdot BC \implies \frac {GD' }{GE' } = \frac {AC}{BC}.$

Point $L$ is the isogonal conjugate of point $G$ with respect to $\triangle ABC \implies \frac {LE}{LD} =\frac {GD' }{GE' } =\frac {AC}{BC}.$

Similarly, one can get $\frac {LE}{AC} = \frac {LD}{BC} = \frac {LF}{AB} = k.$

The double area of $\triangle DLE$ is $2[DLE] = LD \cdot LE \sin \angle DLE = k BC \cdot k AC \cdot \sin \angle ACB = k^2 \cdot 2[ABC].$

Similarly, one can get $[DLE] = [DLF] = [DEF] = k^2 [ABC] \implies L$ is the centroid of $\triangle DEF.$

Corollary

Vector sum $\vec {LE} + \vec {LD} + \vec {LF} = \vec 0.$

Each of these vectors is obtained from the triangle side vectors by rotating by $90^\circ$ and multiplying by a constant $k^2,$ $\vec {AC} + \vec {CB} + \vec {BA} = \vec 0.$

vladimir.shelomovskii@gmail.com, vvsss

Parallel lines

Let $\triangle ABC$ and it’s Lemoine point $L$ be given.

Let $D$ be an arbitrary point. Let $D'$ be the foot from $D$ to line $\overline{AC}$ .

Denote $\ell$ the line through $D$ and parallel to $AC.$

Denote $\ell'$ the line parallel to $AB$ such that distance $EE' = DD' \cdot \frac {AB}{AC}$ and points $E$ and $D$ are both in the exterior (interior) of $\triangle ABC.$

Prove that points $F = \ell \cap \ell', A,$ and $L$ are collinear.

Proof

Denote $P(Q)$ the foot from $L$ to $\overline{AB}(\overline{AC})$ .

$\frac {PL}{AB} = \frac {QL}{AC} \implies \frac {PL}{EE'} = \frac {QL}{DD'}.$ Denote $F = AL \cap \ell, F' = AL \cap \ell' \implies$ $\frac {FA}{AL} = \frac {DD'}{QL} = \frac {EE'}{PL} = \frac {F'A}{AL} \implies F = F'.$

Corollary

If squares $ABGF$ and $ACDE$ are constructed in the exterior of $\triangle ABC,$ then $AO,$ where $O$ is the center of circle $\odot AEF,$ is the symmedian in $\triangle ABC$ through $A.$

vladimir.shelomovskii@gmail.com, vvsss

Radical axis

Circle $\omega$ passes through points $A$ and $B$ and touches line $AC,$ circle $\theta$ passes through points $A$ and $C$ and touches line $AB.$ Let $L$ be the Lemoine point of $\triangle ABC.$

Prove that the radical axis of these circles contains the symmedian of $\triangle ABC.$

Proof

Denote centers of $\omega, \theta,$ and $\odot ABC$ throught $Q,P,$ and $O,$ respectively.

Denote $\ell$ line throught $P$ parallel to $AC, \ell'$ line throught $Q$ parallel to $AB.$ $\angle QAB = |90^\circ - \angle BAC| = \angle PAC \implies \triangle ABQ \sim \triangle ACP \implies$

The ratio of distance from $P$ to $AC$ to $AC$ is equal to the ratio of distance from $Q$ to $AB$ to $AB \implies X = \ell \cap \ell' \in AL.$ $AP \perp AB \implies AP \perp \ell', QA \perp AC \implies QA \perp \ell \implies$

$A$ is the orthocenter of $\triangle PQX \implies XA \perp QP \implies$

the radical axis of these circles contains the $A-$ symmedian of $\triangle ABC.$

Corollary

Circumcenter of $A-$ Apollonius circle point $M,$ circumcenter $O,$ the points on $\odot ABT$ and $\odot ACT$ opposite $A$ belong the line perpendicular $A-$ symmedian of $\triangle ABC.$

vladimir.shelomovskii@gmail.com, vvsss

Construction of symmedian’s point

Let triangle $\triangle ABC$ be given.

Let $D$ be the arbitrary point on sideline $AB, DE||BC, E \in AC.$

The lines $BE$ and $CD$ meet at point $F.$ The circumcircles of triangles $\triangle BFD$ and $\triangle CEF$ meet at two distinct points $F$ and $Q.$

Prove that the line $AQ$ is the $A-$ symmedian of $\triangle ABC.$

Proof

The spiral similarity centered at $Q$ with the angle of rotation $\angle BQE$ maps point $B$ to point $E$ and point $D$ to point $C.$

So $\triangle BDQ \sim \triangle ECQ \implies$

$\angle DBQ = \angle CEQ \implies \angle ABQ + \angle AEQ = 180^\circ \implies ABQE$ is concyclic.

Let $H$ and $G$ be the foot from $Q$ to $\overline{AB}$ and to $\overline{AC},$ respectively.

$\frac {AB}{AC} = \frac {BD}{EC} = \frac {BQ}{QE} = \frac {HQ}{QG}$ which means that $Q$ lies on $A-$ symmedian.

vladimir.shelomovskii@gmail.com, vvsss

Common Lemoine point

Let $\triangle ABC$ be given, $\Omega = \odot ABC.$

Let $L$ be the Lemoine point of $\triangle ABC.$

$A' = AL \cap \Omega \ne A, B' = BL \cap \Omega \ne B, C' = CL \cap \Omega \ne C.$

Prove that the point $L$ is the Lemoine point of $\triangle A'B'C'.$

Proof

Denote point $D$ so that $LD \perp BC, D \in BC.$

Similarly denote $E \in AC$ and $F \in AB.$ $L$ is the centroid of $\triangle DEF.$

$\triangle DEF \sim \triangle A'B'C'$ (see Claim).

Let point $G$ be the centroid of $\triangle A'B'C' \implies$ $\angle LDE = \angle GA'B'.$ $CDLE$ is cyclic so $\angle LDE = \angle LCE = \angle LCA = \angle C'CA = \angle C'A'A = \angle C'A'L$ therefore $A'L$ and $A'G$ are isogonals with respect $\angle C'A'B'.$

Similarly $B'L$ and $B'G$ are isogonals with respect $\angle A'B'C' \implies$

$L$ is the isogonal conjugate of a point $G$ with respect to a triangle $\triangle A'B'C'$

so $L$ is the Lemoine point of $\triangle A'B'C'.$

Claim

Lines AP, BP and CP intersect the circumcircle of $\triangle ABC$ at points $A', B',$ and $C'.$

Points $D, E,$ and $F$ are taken on the lines $BC, CA,$ and $AB$ so that $\angle PDB = \angle PFA = \angle PEC$ (see diagram).

Prove that $\triangle A'B'C' \sim \triangle DEF.$

Proof

$\angle PFA = \angle PDB \implies PDBF$ is cyclic so $\angle PDF = \angle PBF = \angle ABB' = \angle AA'B'.$

Similarly, $\angle PDE = \angle AA'C' \implies$ $\angle FDE = \angle PDF + \angle PDE = \angle AA'B' + \angle AA'C' = \angle B'A'C'.$

Similarly, $\angle DEF = \angle A'B'C'. \blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

Lemoine point extreme properties

Lemoine point $L$ minimizes the sum of the squares of the distances to the sides of the triangle (among all points internal to $\triangle ABC.)$

Proof

Let us denote the desired point by $X.$ Let us imagine that point $X$ is connected to springs of equal stiffness attached to the sides at points $D, E,$ and $F$ and contacts sliding along them without friction. The segments modeling the springs will be perpendicular to the corresponding side. The energy of each spring is proportional to the square of its length. The minimum energy of the system corresponds to the minimum of the sum of the squares of the lengths of these segments, that is, the sum of the squares of the distances from $X$ to the sides.

It is known that the minimum spring energy corresponds to the equilibrium position. The condition of equilibrium at a point $X$ is the equality to zero of the vector sum of forces applied from the springs to the point $X.$ The force developed by each spring is proportional to its length, that is, the equilibrium condition is that the sum of the vectors $\vec {XE} + \vec {XD} + \vec {XF} = \vec 0.$ It is clear that the point $L$ corresponds to this condition.

vladimir.shelomovskii@gmail.com, vvsss

Lemoine point and perpendicularity

Let $\triangle ABC$ be given. Let $L$ be the Lemoine point of $\triangle ABC,$ $LD \perp BC, D \in BC, LE \perp AC, E \in AC, LF \perp AB, F \in AB.$

$M$ is the midpoint $BC.$

Prove that $FE \perp AM.$

Proof

$AL$ is isogonal conjugated $AM$ with respect $\angle A \implies \angle BAL = \angle CAM.$

$LE \perp AE, LF \perp AF \implies AELD$ is cyclic.

$\angle CAM = \angle BAL = \angle FEL.$

$LE \perp AC \implies EF \perp AM.$

vladimir.shelomovskii@gmail.com, vvsss

Lemoine point line

Let $\triangle ABC$ be given. Let $L$ be the Lemoine point of $\triangle ABC.$

Let $AH$ be the height, $AM$ be the median, $LD \perp BC, D \in BC,$

$LE \perp AC, E \in AC, LF \perp AB, F \in AB, P$ be the midpoint $AH$ .

Prove that the points $L, P,$ and $M$ are collinear.

Proof

Denote $O$ the circumcenter $\odot AELF, LO = AO.$

Denote $T$ the midpoint $FE \implies OT \perp FE.$

$L$ is centroid of $\triangle DEF \implies DLT$ is $D-$ median of $\triangle DEF.$

Denote $Q$ the point symmetric $L$ with respect $T \implies QT$ is the midline of $\triangle LAQ \implies AQ \perp EF \implies Q \in AM \perp EF.$

$LD = 2 TL \implies DL = LQ \implies ML$ is the median of $\triangle MDQ.$

$MP$ is the median of $\triangle MHA, HA || DQ \implies$ the points $L, P,$ and $M$ are collinear.

vladimir.shelomovskii@gmail.com, vvsss

Antiparallel lines and segments

Two lines $BC$ and $B'C'$ are said to be antiparallel with respect to the sides of an angle $A$ if they make the same angle in the opposite senses with the bisector of that angle.

A segment $B'C',$ where points $B'$ and $C'$ lie on rays $AC$ and $AB,$ is called antiparallel to side $BC$ if $\angle AB'C' = \angle ABC$ and $\angle AC'B' = \angle ACB.$ The points $B', C', B,$ and $C$ are concyclic.

Prove that the symmedian $AS$ bisects any segment $B'C'$ iff it is antiparallel to side $BC.$

Proof

1) Let segment $B'C'$ be the antiparallel to side $BC.$ Reflection through the bisector of angle $A$ maps the segment $B'C'$ into a segment parallel to side $BC,$ and maps the symmedian $AS$ into the median which bisects image of $B'C'.$

2) Suppose the symmedian $AS$ bisects the segment $DE,D \in AB, D \ne B', E \in AC, E \ne C'$ in point $M.$ There is a segment $B'C'$ with ends on the sides of angle $A$ which contain point $M$ and is antiparallel to side $BC.$ $M$ is the midpoint $B'C' \implies$

$DB'EC'$ is parallelogram, so $DB' || EC'. DB' \in AB, EC' \in AC -$ contradiction.

vladimir.shelomovskii@gmail.com, vvsss

Three intersecting antiparallel segments

Let triangle $ABC$ and a point $L$ lying inside it be given. Let $D'E, DF',$ and $FE'$ be three segments antiparallel to $AB, AC,$ and $BC,$ respectively. $L \in D'E, L \in DF', L \in FE'.$

Prove that $DF' = D'E = E'F$ iff $L$ is a Lemoine point.

Proof

$\angle BAC = \angle BDF' = \angle CD'E \implies LD = LD'.$ Similarly, $LE = LE', LF = LF'.$

1. Let $L$ be the Lemoine point. So $LD = LF', LF = LE', LE = LD' \implies DF' = D'E = E'F = 2 LD.$

2. Let $DF' = D'E = E'F \implies$ $LD + LF' = LD' + LE = LE' + LF \implies$ $LD + LF = LD + LE = LE + LF \implies$ $LD = LE = LF = LD' = LE' = LF' \implies L$ lies on each symmedian.

vladimir.shelomovskii@gmail.com, vvsss

Three intersecting parallel to sides segments

Let triangle $ABC$ be given. $O$ is it’s circumcenter, $L$ is it’s Lemoine point.

Let $A''B', B''C',$ and $C''A'$ be three segments parallel to $AB, BC,$ and $CA,$ respectively. $L \in A''B', L \in B''C', L \in C''A'.$

Prove that points $A', B', C', A'', B'',$ and $C''$ lies on the circle centered at midpoint $LO$ (the first Lemoine circle).

Proof

$AB' || C''L, AC'' || B'L \implies AB'LC''$ is the parallelogram.

Denote $P = AL \cap B'C'', B'P = PC'' \implies B'C''$ is antiparallel to $BC.$

Similarly, $A''C'$ is antiparallel to $AC, A'B''$ is antiparallel to $AB.$ $\triangle AB'C'' \sim \triangle ABC.$ Denote $c = AB, b = AC, a = BC.$

The coefficient of similarity is $k = \frac {AC''}{AC} = \frac {AC''}{AB} \cdot \frac {AB}{AC} = \frac {b^2}{a^2 +b^2+c^2} \cdot \frac {c}{b} = \frac {bc}{a^2 +b^2+c^2}.$

Therefore $B'C'' = k \cdot BC = \frac {abc}{a^2 +b^2+c^2}.$ Similarly, $A'B'' = B'C'' = C'A''.$

Denote $P' = BL \cap A''C', P'' = CL \cap A'B'' \implies AP = PL, BP' = P'L, CP'' = P''L.$

$\triangle PP'P'' \sim \triangle ABC.$ The coefficient of similarity is $\frac{1}{2}.$

Denote $Q$ the midpoint $OL.$ The midline of $\triangle LAO$ is $QP = \frac {OA}{2}$

Similarly, $P'Q = P''Q = PQ \implies Q$ is circumcenter of $\triangle PP'P''.$

$B'C''$ is antiparallel to $BC,$ so $B'C''$ is tangent to $\odot PP'P'', P$ is the midpoint $B'C''.$

Similarly, $A'B'' = B'C'' = C'A''$ are tangent to $\odot PP'P'', P'$ is the midpoint $A''C', P''$ is the midpoint $A'B''.$

Therefore $A'Q = B'Q = C'Q = A''Q = B''Q = C''Q. \blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

Tucker circle

Let triangle $ABC$ be given. $O$ is it’s circumcenter, $L$ is it’s Lemoine point.

Let homothety centered at $L$ with factor $k$ maps $\triangle ABC$ into $\triangle DEF$ .

Denote the crosspoints of sidelines these triangles as $A' = BC \cap DF, B' = AC \cap DE, C' = AB \cap EF,$ $A'' = BC \cap DE, B'' = AC \cap EF, C'' = AB \cap DF.$

Prove that points $A', B', C', A'', B'',$ and $C''$ lies on the circle centered at $LO$ (Tucker circle).

Proof

$AB' || C''D, AC'' || B'D \implies AB'DC''$ is the parallelogram.

Denote $K = AD \cap B'C'', AK = KD, B'K = KC'' \implies$ $B'C''$ is antiparallel to $BC.$

Similarly, $A''C'$ is antiparallel to $AC, A'B''$ is antiparallel to $AB.$

$M = BE \cap A''C'$ is midpoint $BE, N = A'B'' \cap CF$ is the midpoint $CF.$

$\triangle AB'C'' \sim \triangle ABC.$

$\frac {LD}{AL} = k, AL = LD \implies \frac {AK}{AL} = \frac{2}{k+1}, \frac {KL}{AL} = \frac{1-k}{1+k}.$

Similarly, $\frac {BM}{BL} = \frac {CN}{CL} = \frac {AK}{AL} =\frac{2}{k+1}, \frac {KL}{AL} = \frac{ML}{BM}= \frac {NL}{CL} = \frac{1-k}{1+k}.$

Let $B'''C'''$ be the symmedian $BC$ through $L.$ $B'''C''' || B'C'' \implies B'C'' = B'''C''' \cdot \frac {AK}{AL} =\frac{2B'''C'''}{k+1}.$

It is known that three symmedians through $L$ are equal, so $A''C' = C''B' = B''A' = \frac{2B'''C'''}{k+1}.$

$\triangle KMN$ is homothetic to $\triangle ABC$ with center $L$ and factor $\frac{1-k}{1+k}.$

So segments $A''C' = C''B' = B''A'$ are tangents to $\odot KMN$ and points of contact are the midpoints of these segments.

Denote $Q$ the circumcenter of $\triangle KMN, Q \in LO.$

Therefore $A'Q = B'Q = C'Q = A''Q = B''Q = C''Q. \blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

Tucker circle 2

Let triangle $ABC$ be given. Let $D$ be the arbitrary point on sideline $AC.$

Let $DD'$ be the antiparallel to side $AB, D' \in BC.$

Denote point $E \in AB, D'E || AC.$

Let $EE'$ be the antiparallel to side $BC, E' \in AC.$

Denote point $F \in BC, E'F || AB.$

Let $FF'$ be the antiparallel to side $AC, F' \in AB.$

Prove that points $D, D', E, E', F,$ and $F'$ lies on the circle centered at $LO$ (Tucker circle).

Proof

$\angle ABC = \angle CDD' = \angle AE'E, DE' || DE' \implies DD'EE'$ is isosceles trapezoid.

So $DD' = EE'.$

$\angle ACB = \angle BF'F = \angle AEE', FE' || EF' \implies EFE'F'$ is isosceles trapezoid.

So $FF' = EE' = DD'. \angle BFF' = \angle CD'D \implies F'D || BC.$

Denote $A'$ the midpoint $EE', B'$ the midpoint $FF', C'$ the midpoint $DD'.$ $AB || FE' \implies A'B' ||AB.$ Similarly, $A'C' ||AC, C'B' ||CB.$

$A'$ is the midpoint of antiparallel of $BC \implies AA'$ is the $A-$ symmedian of $\triangle ABC.$

Similarly, $BB'$ is the $B-$ symmedian, $CC'$ is the $C-$ symmedian of $\triangle ABC.$

Therefore Lemoine point $L = AA' \cap BB' \cap CC', \triangle A'B'C'$ is homothetic to $\triangle ABC$ with center $L.$

So segments $DD' = EE' = FF'$ are tangents to $\odot A'B'C'$ and points of contact are the midpoints of these segments.

Denote $Q$ the circumcenter of $\triangle A'B'C', Q \in LO,$ where $O$ is the circumcenter of $\triangle ABC.$

Therefore $DQ = D'Q = EQ = E'Q = FQ = F'Q. \blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

@@ Line 156: / Line 156: @@
 Circumcenter of <math>A-</math>Apollonius circle point <math>M,</math> circumcenter <math>O,</math> the points on <math>\odot ABT</math> and <math>\odot ACT</math> opposite <math>A</math> belong the line perpendicular <math>A-</math>symmedian of <math>\triangle ABC.</math>
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Construction of symmedian’s point==
+[[File:Symmedian from parallel.png|390px|right]]
+Let triangle <math>\triangle ABC</math> be given.
+Let <math>D</math> be the arbitrary point on sideline <math>AB, DE||BC, E \in AC.</math>
+The lines <math>BE</math> and <math>CD</math> meet at point <math>F.</math> The circumcircles of triangles <math>\triangle BFD</math> and <math>\triangle CEF</math> meet at two distinct points <math>F</math> and <math>Q.</math>
+Prove that the line <math>AQ</math> is the <math>A-</math>symmedian of <math>\triangle ABC.</math>
+<i><b>Proof</b></i>
+The spiral similarity centered at <math>Q</math> with the angle of rotation <math>\angle BQE</math> maps point <math>B</math> to point <math>E</math> and point <math>D</math> to point <math>C.</math>
+So <math>\triangle BDQ \sim \triangle ECQ \implies </math>
+<math>\angle DBQ = \angle CEQ \implies \angle ABQ + \angle AEQ = 180^\circ \implies ABQE</math> is concyclic.
+Let <math>H</math> and <math>G</math> be the foot from <math>Q</math> to <math>\overline{AB}</math> and to <math>\overline{AC},</math> respectively.
+<math>\frac {AB}{AC} = \frac {BD}{EC} =  \frac {BQ}{QE} =  \frac {HQ}{QG}</math> which means that <math>Q</math> lies on <math>A-</math>symmedian.
 '''vladimir.shelomovskii@gmail.com, vvsss'''

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Difference between revisions of "Symmedians, Lemoine point"

Revision as of 14:46, 4 August 2024

Contents

Proportions

Radical axis of circumcircle and Apollonius circle

Symmedian and tangents

Lemoine point properties

Parallel lines

Radical axis

Construction of symmedian’s point

Common Lemoine point

Lemoine point extreme properties

Lemoine point and perpendicularity

Lemoine point line

Antiparallel lines and segments

Three intersecting antiparallel segments

Three intersecting parallel to sides segments

Tucker circle

Tucker circle 2