Let the point of intersection of and be , where and are the Brocard Points, and similarly define and . is the first Brocard triangle, and is inversely similar to . It is inscribed in the Brocard Circle drawn with as the Diameter. The triangles , , and are Isosceles Triangles with base angles , where is the Brocard Angle. The sum of the areas of the Isosceles Triangles is , the Area of Triangle . The first Brocard triangle is in perspective with the given Triangle, with , , and Concurrent. The Median Point of the first Brocard triangle is the Median Point of the original triangle. The Brocard triangles are in perspective at .

Let , , and and , , and be the Circles intersecting in the Brocard Points and , respectively. Let the two circles and tangent at to and , and passing respectively through and , meet again at . The triangle is the second Brocard triangle. Each Vertex of the second Brocard triangle lies on the second Brocard Circle.

The two Brocard triangles are in perspective at .

**References**

Johnson, R. A. *Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle.* Boston, MA:
Houghton Mifflin, pp. 277-281, 1929.

© 1996-9

1999-05-26