Difference between revisions of "Symmedians, Lemoine point"
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Then <math>BE</math> is the <math>B-</math> symmedian of <math>\triangle ABD, CE</math> is the <math>C-</math> symmedian of <math>\triangle ACD, DE</math> is the <math>D-</math> symmedian of <math>\triangle BCD.</math> | Then <math>BE</math> is the <math>B-</math> symmedian of <math>\triangle ABD, CE</math> is the <math>C-</math> symmedian of <math>\triangle ACD, DE</math> is the <math>D-</math> symmedian of <math>\triangle BCD.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Symmedian and tangents== | ||
+ | [[File:Tangents and symmedian.png|390px|right]] | ||
+ | Let <math>\triangle ABC</math> and it’s circumcircle <math>\Omega</math> be given. Tangents to <math>\Omega</math> at points <math>B</math> and <math>C</math> intersect at point <math>F.</math> | ||
+ | Prove that <math>AF</math> is <math>A-</math> symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>D = AF \cap \Omega \ne A.</math> WLOG, <math>\angle BAC < 180^\circ.</math> | ||
+ | <cmath>\triangle FDB \sim \triangle FBA \implies \frac {BD}{AB} = \frac{DF}{BF}.</cmath> | ||
+ | <cmath>\triangle FDC \sim \triangle FCA \implies \frac {CD}{AC} = \frac{DF}{CF}.</cmath> | ||
+ | <math>BF = CF \implies \frac {BD}{CD} = \frac{AB}{AC} \implies AD</math> is <math>A-</math> symmedian of <math>\triangle ABC.</math> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 14:03, 9 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 is isogonally conjugate to the median
There are three symmedians. They are meet at a triangle center called the Lemoine point.
Proportions
Let be given.
Let be the median,
Prove that iff is the symmedian than
Proof
1. Let be the symmedian. So Similarly
By applying the Law of Sines we get Similarly,
2.
As point moves along the fixed arc from to , the function monotonically increases from zero to infinity. This means that there is exactly one point at which the condition is satisfied. In this case, point lies on the symmedian.
Similarly for point
Corollary
Let be the symmedian of
Then is the symmedian of is the symmedian of is the symmedian of
vladimir.shelomovskii@gmail.com, vvsss
Symmedian and tangents
Let and it’s circumcircle be given. Tangents to at points and intersect at point Prove that is symmedian of
Proof
Denote WLOG, is symmedian of
vladimir.shelomovskii@gmail.com, vvsss