Isogonal Conjugates Proof of Existence (Symmedians)
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 moves when you move . is the incenter.
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.
I recommend checking the Brilliant wiki for additional resources.
Partial thanks to https://www.math.cmu.edu/~ploh/docs/math/mop2008/collin-concur-soln.pdf for inspiration (problem 2 using isogonal conjugate of Gergonne point). I came up with the rest of the proof, and any plagiarism is accidental.
By the converse of Ceva's theorem, exists iff .
Imagine that we want to flip and scale it down so that , , and are all similar to , but reversed so that the lines that were reflected across the angle bisectors now coiincide at in the smaller triangles. This would imply that , so not only are triangles and similar, triangles and are as well. So . Using the same reasoning, and . However, so all angles are 90.
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.
https://brilliant.org/wiki/triangles-orthocenter/
and in the diagram represent the orthocenter and incenter respectively. This makes , , and the feet of the altitudes from , , and respectively. Point is arbitrary.
Let be the intersection of and , be the intersection of and , and be the intersection of and .
Notice that since is similar to by reflection over the angle bisector , and reflecting over the angle bisector would land on , splits the line in the same ratio that does. Same goes for and . So we have translated the problem into exists iff .
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, , and . So . However is a straight line( is an intersection). So . When we combine this for each side we get . Shifting the demoninators over one yields . Since they have the same bases, , , and . So, ultimately, . But , , and 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)