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- ...o]]s hold among these points. In particular, <math>\overline{OGNH}</math> and <math>OG:GN:NH = 2:1:3</math> ...er lines of <math>\triangle AH_BH_C</math>,<math>\triangle BH_CH_A</math>, and <math>\triangle CH_AH_B</math> [[concurrence | concur]] at <math>N</math>,59 KB (10,203 words) - 03:47, 30 August 2023
- ...</math> with respect to the lines containing the bisectors <math>AI</math> and <math>CI,</math> respectively. ...<math>F</math> at which the segments of the bisectors <math>AI, BI,</math> and <math>CI,</math> respectively intersect the incircle of <math>\triangle ABC19 KB (3,291 words) - 12:44, 6 October 2024
- ...one of the <math>XY</math> arcs of <math>\Gamma</math>. Let <math>A</math> and <math>B</math> be two points on the segment <math>XY</math>. The straight l ...and <math>NB</math> cut <math>\Gamma</math> again at points <math>C</math> and <math>D</math>, respectively. The tangents to <math>\Gamma</math> in2 KB (306 words) - 10:19, 5 May 2024
- The Tucker circles are a generalization of the cosine circle and first Lemoine circle. Prove that points <math>A', B', C', A'', B'',</math> and <math>C''</math> lies on the circle centered at <math>LO</math> (Tucker cir4 KB (731 words) - 05:34, 4 August 2024
- Let two planes <math>\Pi</math> and <math>\Pi'</math> and a point <math>O</math> not lying in them be defined in space. To each point ...lowing problem. Let us construct a plane containing a point <math>O</math> and parallel to the plane <math>\Pi'.</math> Let us denote the line along which39 KB (6,959 words) - 14:26, 3 December 2024
- ...the angle bisector, but it is on the other side of the angle bisector. The symmedian <math>AS_A</math> is isogonally conjugate to the median <math>AM_A.</math> Prove that iff <math>AE</math> is the symmedian than <math>\frac {BD}{CD} = \frac{AB}{AC}, \frac {BE}{CE} = \left (\frac{AB30 KB (5,152 words) - 00:30, 2 January 2025