A point D in a triangle ABC such that $\angle DAB= \angle DBC= \angle DCA$ I got this question from a student of mine, who is participating in a math olympiad competition: How can we construct a point D in a triangle ABC such that $\angle DAB= \angle DBC= \angle DCA$? I've tried with the most common points such as barycenter or orthocenter, but nothing. I appreciate any hint. Thanks in advance.
 A: You are asking for so called Brocard's point of the triangle. One of constructions can be found on Wikipedia: http://en.m.wikipedia.org/wiki/Brocard_points
A: draw a line $l$ at $A$ that is perpendicular to $AB.$  draw perpendicular bisector $m$ of the side $AC.$  let the lines cut at $B'.$  now draw a circle $\beta$ with center $B'$ and radius $B'A = B'C.$ the circle $\beta$ is tangent to the side $AB$ and goes through $C.$ in addition, if $D$ is point inside the triangle $ABC$ and on the circle $\beta,$  then $\angle BAD= \angle ACD.$ 
repeat the constructions at the other two sides and get circles $\alpha$ through $B, c$ and circle $\gamma$ through  $A, B.$
the point $D$ is the intersection of these three circles. in fact this point has a name that i am unable to recall.
A: Construct an equilateral triangle on the outside of two of the edges and find its midpoint. Use this to draw an arc through the two associated vertices of your original triangle. Repeat this for another side of the triangle ABC. The required point D is at the intersection of these two arcs.

