Isogonal conjugate
Isogonal conjugates are pairs of points in the plane with respect to a certain triangle.
Construction and Theorem
Let 
be a point in the plane, and let 
be a triangle. We will denote by 
the lines 
. Let 
denote the lines 
, 
, 
, respectively. Let 
, 
, 
be the reflections of 
, 
, 
over the angle bisectors of angles 
, 
, 
, respectively. Then lines , , concur at a point 
, called the isogonal conjugate of with respect to triangle .
Proof
By our constructions of the lines 
, 
, and this statement remains true after permuting . Therefore by the trigonometric form of Ceva's Theorem
![\[\frac{\sin \angle q_a b}{\sin \angle c q_a} \cdot \frac{\sin \angle q_b c}{\sin \angle a q_b} \cdot \frac{\sin \angle q_c a}{\sin \angle b q_c} = \frac{\sin \angle p_a c}{\sin \angle b p_a} \cdot \frac{\sin \angle p_b a}{\sin \angle c p_b} \cdot \frac{\sin \angle p_c b}{\sin \angle a p_c} = 1,\]](http://latex.artofproblemsolving.com/9/f/8/9f814834eac4f9820c758827b981c84a7950ab99.png)
so again by the trigonometric form of Ceva, the lines 
concur, as was to be proven. 
Problems
Olympiad
Given a nonisosceles, nonright triangle 
let 
denote the center of its circumscribed circle, and let 
and 
be the midpoints of sides 
and 
respectively. Point 
is located on the ray 
so that 
is similar to 
. Points 
and 
on rays 
and 
respectively, are defined similarly. Prove that lines 
and 
are concurrent, i.e. these three lines intersect at a point. (Source)
Let be a given point inside quadrilateral 
. Points 
and 
are located within such that 
, 
, 
, 
. Prove that 
if and only if 
. (Source)
