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Difference between revisions of "Symmedians, Lemoine point"

(→‎Symmedian and tangents)
(→‎Lemoine point properties)
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Similarly, one can get <math>[DLE]  = [DLF] = [DEF] = k^2 [ABC] \implies L</math> is the centroid of <math>\triangle DEF.</math>
 
Similarly, one can get <math>[DLE]  = [DLF] = [DEF] = k^2 [ABC] \implies L</math> is the centroid of <math>\triangle DEF.</math>
 +
 +
<i><b>Corollary</b></i>
 +
 +
Vector sum <math>\vec {LE} + \vec {LD} + \vec {LF} = \vec 0.</math>
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Each of these vectors is obtained from the triangle side vectors by rotating by <math>90^\circ</math> and multiplying by a constant <math>k^2,</math>
 +
<cmath>\vec {AC} + \vec {CB} + \vec {BA} = \vec 0.</cmath>
  
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 05:09, 20 July 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.

Proportions

Symedian segments.png

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

Symmedian and tangents

Tangents and symmedian.png

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

Tangents to symmedian.png

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

L and G.png

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

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