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symmedian
symmedian, n. and a. Geom. (sɪˈmiːdɪən) [f. Gr. σύν sym- + median a.1 and n.1] symmedian, or symmedian line, each of three lines drawn from the angles of a triangle at inclinations to the angle-bisectors equal to those of the medians (i.e. the lines from the angles to the middle points of the opposi...
Oxford English Dictionary
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Symmedian
The symmedian planes can be shown to intersect at a point, the symmedian point. References
External links
Symmedian and Antiparallel at cut-the-knot
Symmedian and 2 Antiparallels at cut-the-knot
Symmedian and the Tangents at cut-the-knot
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cosymmedian
cosymmedian, a. Math. (kəʊsɪˈmiːdɪən) [f. co- 2 + symmedian.] Of triangles: Having the same symmedian lines.1888 J. J. Milne Companion to Weekly Problem Papers 147 Triangles ABC, A{p}B{p}C{p} so related, and having the same symmedian lines AKA{p}, BKB{p}, CKC{p}, are called Cosymmedian triangles. Ib...
Oxford English Dictionary
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Exsymmedian
For a triangle with being the exsymmedians and being the symmedians through the vertices , two exsymmedians and one symmedian intersect in a common
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Mittenpunkt
It was identified in 1836 by Christian Heinrich von Nagel as the symmedian point of the excentral triangle of the given triangle. Collinearities
The mittenpunkt is at the intersection of the line connecting the centroid and the Gergonne point, the line connecting the incenter and the symmedian
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How to prove the property of the Lemoine point of a triangle? From Wolfram MathWorld, I know there is a Lemoine point of triangle, also called symmedian point, the sum of squared distances of this point to all the thr...
Let $(d_a,d_b,d_c)$ be the distances of your point from the sides. You have to minimize the quantity: $$ d_a^2+d_b^2+d_c^2 $$ under the constraint: $$ a d_a + b d_b + c d_c = 2\Delta,$$ hence Lagrange multipliers gives that $(d_a,d_b,d_c)=\lambda(a,b,c)$, so your stationary point is the isogonal con...
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Lemoine hexagon
hexagon with vertices given by the six intersections of the edges of a triangle and the three lines that are parallel to the edges that pass through its symmedian The second is a self-intersecting hexagon with the lines going through the symmedian point as three of the edges and the other three edges join pairs of
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Lemoine point
In the Encyclopedia of Triangle Centers the symmedian point appears as the sixth point, X(6). Line is a symmedian, as can be seen by drawing the circle with center through and .
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show that $PY=PB$ is such $\Delta YAC$ is equilateral If P is a point on $B$ median of $ABC$ with $\angle B=60$,Let $X$ be the reflection $B$ wrt midpt of $AC$,and $X=P$,Let $Y$ be a point on the same side of $AC$ as ...
With a couple of additional points is pretty trivial. $P'$ and $C'$ are chosen in such a way that both $CBC'$ and $PAP'$ are equilateral.  the symmedian point .
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Trilinear polarity
The trilinear polar of the symmedian point is the Lemoine axis of triangle .
The trilinear polar of the orthocenter is the orthic axis. For example, the Trilinear polar of a point on the circumcircle must pass through its perspector, the Symmedian point X(6).
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Brocard circle
It passes through the circumcenter and symmedian of the triangle, and is centered at the midpoint of the line segment joining them (so that this segment These five points, together with the other two points on the circle (the circumcenter and symmedian), justify the name "seven-point circle".
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Tangential triangle
The center of perspectivity, where these three lines meet, is the symmedian point of the triangle. Any two of these are concurrent with the third symmedian of the reference triangle.
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