Circles (A42a) and (A42b): QTB circles


Let T be the point on CD beyond D such that CT = AB. Let H be the orthocenter of ATB, also the highest point of Bankoff's triplet circle (A2). Let A42a be the midpoint of AH, then A42a lies on the Nine-Point Circle of ATB. The circle with center A42a through O1 is Archimedean. Similarly we find A42b on BH, and the Archimedean circle with center A42b through O2. These circles are named after their author Quang Tuan Bui. [Bui 2007]

Back to Catalogue