A "symmedian" is a reflection of the median of a triangle across an angle bisector. Interestingly, all three symmedians of a triangle meet at a point. However, this fact is not unique to medians. Any point has an associated isogonal conjugate.(denoted S in the activity). If this confuses you, focus on how [math]D_1[/math] moves when you move [math]D[/math]. [math]I[/math] is the incenter.[br][br]A proof is below. You need to know Ceva's theorem in order to follow it.

Note that one can trivially prove the existence of isogonal conjugates by using trig ceva.[br]I recommend checking the Brilliant wiki for additional resources.[br]Partial thanks to [url=https://www.math.cmu.edu/~ploh/docs/math/mop2008/collin-concur-soln.pdf]https://www.math.cmu.edu/~ploh/docs/math/mop2008/collin-concur-soln.pdf[/url] for inspiration (problem 2 using isogonal conjugate of Gergonne point). I came up with the rest of the proof, and any plagiarism is accidental. [br][br]By the converse of Ceva's theorem, [math]S[/math] exists iff [math]\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1[/math].[br][br]Imagine that we want to flip [math]ABC[/math] and scale it down so that [math]AFG[/math], [math]BGE[/math], and [math]CEF[/math] are all similar to [math]ABC[/math], but reversed so that the lines that were reflected across the angle bisectors now coiincide at [math]D[/math] in the smaller triangles. This would imply that [math]\frac{AF}{AG}=\frac{AB}{AC}\Longrightarrow\cdot\frac{AF}{AB}=\frac{AG}{AC}[/math], so not only are triangles [math]AFG[/math] and [math]ABC[/math] similar, triangles [math]AFB[/math] and [math]AGC[/math] are as well. So [math]\angle AFB=\angle AGC[/math]. Using the same reasoning, [math]\angle BEA=\angle BGC[/math] and [math]\angle CEA=\angle CFB[/math]. However, [math]\angle BGC=180-\angle AGC\wedge\angle BEA=180-\angle CEA\Longrightarrow AFB=CFB[/math] so all angles are 90.[br]I knew this fact beforehand, so it's understandable if you feel it came out the blue. Just while we're at it, if you don't already know a proof for the existence of the orthocenter, please read it as it will be useful later on.[br][url=https://brilliant.org/wiki/triangles-orthocenter/]https://brilliant.org/wiki/triangles-orthocenter/[br][br][math]O[/math][/url] and [math]I[/math] in the diagram represent the orthocenter and incenter respectively. This makes [math]E[/math], [math]F[/math], and [math]G[/math] the feet of the altitudes from [math]A[/math], [math]B[/math], and [math]C[/math] respectively. Point [math]D[/math] is arbitrary.[br][br]Let [math]L[/math] be the intersection of [math]AD[/math] and [math]FG[/math], [math]M[/math] be the intersection of [math]BD[/math] and [math]GE[/math], and [math]N[/math] be the intersection of [math]CD[/math] and [math]EF[/math].[br][br]Notice that since [math]AFG[/math] is similar to [math]ABC[/math] by reflection over the angle bisector [math]AI[/math], and reflecting [math]X[/math] over the angle bisector [math]AI[/math] would land [math]X[/math] on [math]AD[/math], [math]L[/math] splits the line in the same ratio that [math]X[/math] does. Same goes for [math]M[/math] and [math]N[/math]. So we have translated the problem into [math]S[/math] exists iff [math]\frac{EN}{NF}\cdot\frac{FL}{LG}\cdot\frac{GM}{ME}=1[/math].[br]Recall that if two triangles have the same altitude and their bases lie on the same line the ratio of their areas are the ratio of their bases. Therefore, [math]\frac{AFL}{AGL}=\frac{FL}{GL}[/math], and [math]\frac{FLD}{GLD}=\frac{FL}{GL}[/math]. So [math]\frac{FALD}{GALD}=\frac{FL}{GL}[/math]. However [math]ALD[/math] is a straight line([math]L[/math] is an intersection). So [math]\frac{FAD}{GAD}=\frac{FL}{GL}[/math]. When we combine this for each side we get [math]\frac{ECD}{FCD}\cdot\frac{FAD}{GAD}\cdot\frac{GBD}{EBD}=\frac{EN}{NF}\cdot\frac{FL}{LG}\cdot\frac{GM}{ME}=\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}[/math]. Shifting the demoninators over one yields [math]\frac{ECD}{EBD}\cdot\frac{FAD}{FCD}\cdot\frac{GBD}{GAD}[/math]. Since they have the same bases, [math]\frac{ECD}{EBD}=\frac{EC}{EB}[/math], [math]\frac{FAD}{FCD}=\frac{FA}{FC}[/math], and [math]\frac{GBD}{GAD}=\frac{GB}{GA}[/math]. So, ultimately, [math]\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{CE}{EB}\cdot\frac{BG}{GA}\cdot\frac{AF}{FC}[/math]. But [math]E[/math], [math]F[/math], and [math]G[/math] form the orthocenter when connected to their opposite points, so the RHS is equal to 1, meaning the isogonal conjugate does exist! (as a result of ceva's theorem)